今莫陪我，正覺無聊甚，遂開了A + B problem。欲久，竟有多至二十一也。今與眾分，共濟美之一日.

1.普通

#include<iostream>
using namespace std;
int main()
{
	int a,b;
	cin >> a >> b;
	cout << a + b;
	return 0;
}

2.高精

#include<iostream>
#include<cstring>
using namespace std;
int main()
{
    char a1[1000],b1[1000];
      int a[1000]={0},b[1000]={0},c[1000]={0},la,lb,lc,i,x;
      cin>>a1>>b1;
      la=strlen(a1);
      lb=strlen(b1);
      for(i=0;i<=la-1;i++){a[la-i]=a1[i]-48;}
    for(i=0;i<=lb-1;i++){b[lb-i]=b1[i]-48;}
      lc=1,x=0;
    while(lc<=la||lc<=lb){c[lc]=a[lc]+b[lc]+x,x=c[lc]/10,c[lc]%=10,lc++;}
    c[lc]=x;
    if(c[lc]==0){lc--;}
    for(i=lc;i>=1;i--){cout<<c[i];}
    cout<<endl;
    return 0;
}
 

3.壓位高精

#include <cstdio>  
#include <cstring>  
#include <cstdlib>  
#include <iostream>  
#define p 8
#define carry 100000000
using namespace std;  
const int Maxn=50001;  
char s1[Maxn],s2[Maxn];  
int a[Maxn],b[Maxn],ans[Maxn];  
int change(char s[],int n[])   
{  
    char temp[Maxn];   
    int len=strlen(s+1),cur=0;  
    while(len/p)
    {  
        strncpy(temp,s+len-p+1,p);
        n[++cur]=atoi(temp); 
        len-=p;
    }  
    if(len)
    {
        memset(temp,0,sizeof(temp));  
        strncpy(temp,s+1,len);  
        n[++cur]=atoi(temp);   
    }  
    return cur;
}  
int add(int a[],int b[],int c[],int l1,int l2)  
{  
    int x=0,l3=max(l1,l2);  
    for(int i=1;i<=l3;i++)
    {  
        c[i]=a[i]+b[i]+x;  
        x=c[i]/carry;
        c[i]%=carry;  
    }  
    while(x>0){c[++l3]=x%10;x/=10;}  
    return l3;
}  
void print(int a[],int len)  
{   
    printf("%d",a[len]);
    for(int i=len-1;i>=1;i--)printf("%0*d",p,a[i]);
    printf("\n");  
}  
int main()  
{
    scanf("%s%s",s1+1,s2+1);
    int la=change(s1,a);
    int lb=change(s2,b);
    int len=add(a,b,ans,la,lb);    
    print(ans,len);
}
 

4.位運算

#include <iostream>
using namespace std;
int plus(int a,int b)//這個是加法運算函式
{
    if(b==0)//如果b(進位)是0(沒有進位了)，返回a的值
        return a;
    else
    {
        int xor,carry;
        xor=a^b;//xor是a和b不進位加法的值
        carry=(a&b)<<1;//carry是a和b進位的值(只有兩個都是1才會產生進位，所以是與運算。左移一位是因為二進位制加法和十進位制加法豎式一樣進位要加在左面一位裡)
        return plus(xor,carry);//把不進位加法和進位的值的和就是結果
    }
}
int main()
{
    int a,b;
    cin >> a >> b;
    cout << plus(a,b) << endl;
    return 0;
}
 

5.線段樹

#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<iostream>
using namespace std;
struct node{
    int val,l,r;
};
node t[5];
int a[5],f[5];
int n,m;
void init(){
    for(int i=1;i<=2;i++){
        scanf("%d",&a[i]);
    }
}
void build(int l,int r,int node){//這是棵樹
    t[node].l=l;t[node].r=r;t[node].val=0;
    if(l==r){
        f[l]=node;
        t[node].val=a[l];
        return;
    }
    int mid=(l+r)>>1;
    build(l,mid,node*2);
    build(mid+1,r,node*2+1);
    t[node].val=t[node*2].val+t[node*2+1].val;
}
void update(int node){
    if(node==1)return;
    int fa=node>>1;
    t[fa].val=t[fa*2].val+t[fa*2+1].val;
    update(fa);
}
int find(int l,int r,int node){
    if(t[node].l==l&&t[node].r==r){
        return t[node].val;
    }
    int sum=0;
    int lc=node*2;int rc=lc+1;
    if(t[lc].r>=l){
        if(t[lc].r>=r){
            sum+=find(l,r,lc);
        }
        else{
            sum+=find(l,t[lc].r,lc);
        }
    }
    if(t[rc].l<=r){
        if(t[rc].l<=l){
            sum+=find(l,r,rc);
        }
        else{
            sum+=find(t[rc].l,r,rc);
        }
    }
    return sum;
}
int main(){
    init();
    build(1,2,1);
    printf("%d",find(1,2,1));
}

6.字典樹

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
struct node{
    int str[26];
    int sum;
}s[1000];
char str1[100];
int t=0,tot=0,ss=0;
bool f1;
void built()
{
    t=0;
    for(int i=0;i<strlen(str1);i++)
    {
         if(str1[i]=='-'){
             f1=true;continue;
         }
         if(!s[t].str[str1[i]-'0'])
         s[t].str[str1[i]-'0']=++tot;
         t=s[t].str[str1[i]-'0'];
         s[t].sum=str1[i]-'0';
    }
}
int query()
{
   int t=0;int s1=0;
   for(int i=0;i<strlen(str1);i++)
   {
           if(str1[i]=='-') continue;
           if(!s[t].str[str1[i]-'0']) return s1;
           t=s[t].str[str1[i]-'0'];
           s1=s1*10+s[t].sum;
   }
   return s1;
}
int main()
{    
  for(int i=1;i<=2;i++)
  {
      f1=false;
      scanf("%s",str1);
    built();
    if(f1)
      ss-=query();
      else ss+=query();
  }
  printf("%d",ss);
  return 0;    
}

7.樹狀陣列

#include<iostream>
#include<cstring>
using namespace std;
int lowbit(int a)
{
    return a&(-a);
}
int main()
{
    int n=2,m=1;
    int ans[m+1];
    int a[n+1],c[n+1],s[n+1];
    int o=0;
    memset(c,0,sizeof(c));
    s[0]=0;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        s[i]=s[i-1]+a[i];
        c[i]=s[i]-s[i-lowbit(i)];//樹狀陣列建立字首和優化
    }
    for(int i=1;i<=m;i++)
    {
        int q=2;
        if(q==1)
        {（沒有更改操作）
            int x,y;
            cin>>x>>y;
            int j=x;
            while(j<=n)
            {
                c[j]+=y;
                j+=lowbit(j);
            }
        }
        else
        {
            int x=1,y=2;//求a[1]+a[2]的和
            int s1=0,s2=0,p=x-1;
            while(p>0)
            {
                s1+=c[p];
                p-=lowbit(p);//樹狀陣列求和操作，用兩個字首和相減得到區間和
            }
            p=y;
            while(p>0)
            {
                s2+=c[p];
                p-=lowbit(p);
            }    
            o++;
            ans[o]=s2-s1;//儲存答案
        }
    }
    for(int i=1;i<=o;i++)
        cout<<ans[i]<<endl;//輸出
    return 0;
}

8.最小生成樹

#include <cstdio>
#include <algorithm>
#define INF 2140000000
using namespace std;
struct tree{int x,y,t;}a[10];
bool cmp(const tree&a,const tree&b){return a.t<b.t;}
int f[11],i,j,k,n,m,x,y,t,ans;
int root(int x){if (f[x]==x) return x;f[x]=root(f[x]);return f[x];}
int main(){
    for (i=1;i<=10;i++) f[i]=i;
    for (i=1;i<=2;i++){
        scanf("%d",&a[i].t);
        a[i].x=i+1;a[i].y=1;k++;
    }
    a[++k].x=1;a[k].y=3,a[k].t=INF;
    sort(a+1,a+1+k,cmp);
    for (i=1;i<=k;i++){
    //    printf("%d %d %d %d\n",k,a[i].x,a[i].y,a[i].t);
        x=root(a[i].x);y=root(a[i].y);
        if (x!=y) f[x]=y,ans+=a[i].t; 
    }
    printf("%d\n",ans);
}

9.二分

#include<cstdio>
using namespace std;
int a,b,c;
int main()
{
	long long l = -int(1e9) << 1,r = int(1e9) << 1;
    scanf("%d%d",&a,&b);
    while(r - l > 1)
	{
		c = (l + r) >> 1;
        if(c - b < a)
			l = c;
        else if(c - b > a)
			r = c;
        else
			return printf("%d\n",c),0;
    }
	if(l != r)
		return printf("%d\n",r),0;
}

10.二進位制

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
int main()
{
    int a,b,s=0,s1=0,i=0,na=0,nb=0;
    cin>>a>>b;
    if(a<=0) na=1,a*=-1;
    while(a!=0)
    {
        if(a%2!=0)
        s+=pow(2,a%2*i);
        a/=2;
        i++;
    }
    i=0;
    if(na==1) s*=-1;
    if(b<=0) nb=1,b*=-1;
    while(b!=0)
    {
        if(b%2!=0)
        s1+=pow(2,b%2*i);
        b/=2;
        i++;
    }
    if(nb==1) s1*=-1;
    cout<<s+s1;;
    return 0;
}

11.讀入&輸出優化

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
int read(){
    int out=0,fh=1;
    char cc=getchar();
    if (cc=='-') fh=-1;
    while (cc>'9'||cc<'0') cc=getchar();
    while (cc>='0'&&cc<='9') {out=out*10+cc-'0';cc=getchar();}
    return out*fh;
}
void write(int x)  
{  
    if (x==0){
        putchar('0');
        return;
    }
    int num = 0; char c[15];
    while(x) c[++num] = (x%10)+48, x /= 10;
    while(num) putchar(c[num--]);
    putchar(' '); 
}
int main(){
    write(read()+read());
    return 0;
}

12.遞迴

#include<iostream>
using namespace std;
long long a,b,c;
long long dg(long long a)
{
    if(a<=5){return a;}
    return (dg(a/2)+dg(a-a/2));
}
int main()
{
    cin>>a>>b;
    c=dg(a)+dg(b);
    cout<<c;
}

13.Splay

#include <bits/stdc++.h>
#define ll long long
#define N 100000
using namespace std;
int sz[N], rev[N], tag[N], sum[N], ch[N][2], fa[N], val[N];
int n, m, rt, x;
void push_up(int x){
    sz[x] = sz[ch[x][0]] + sz[ch[x][1]] + 1;
    sum[x] = sum[ch[x][1]] + sum[ch[x][0]] + val[x];
}
void push_down(int x){
    if(rev[x]){
        swap(ch[x][0], ch[x][1]);
        if(ch[x][1]) rev[ch[x][1]] ^= 1;
        if(ch[x][0]) rev[ch[x][0]] ^= 1;
        rev[x] = 0;
    }
    if(tag[x]){
        if(ch[x][1]) tag[ch[x][1]] += tag[x], sum[ch[x][1]] += tag[x];
        if(ch[x][0]) tag[ch[x][0]] += tag[x], sum[ch[x][0]] += tag[x];
        tag[x] = 0;
    }
}
void rotate(int x, int &k){
    int y = fa[x], z = fa[fa[x]];
    int kind = ch[y][1] == x;
    if(y == k) k = x;
    else ch[z][ch[z][1]==y] = x;
    fa[x] = z; fa[y] = x; fa[ch[x][!kind]] = y;
    ch[y][kind] = ch[x][!kind]; ch[x][!kind] = y;
    push_up(y); push_up(x);
}
void splay(int x, int &k){
    while(x != k){
        int y = fa[x], z = fa[fa[x]];
        if(y != k) if(ch[y][1] == x ^ ch[z][1] == y) rotate(x, k);
        else rotate(y, k);
        rotate(x, k);
    }
}
int kth(int x, int k){
    push_down(x);
    int r = sz[ch[x][0]]+1;
    if(k == r) return x;
    if(k < r) return kth(ch[x][0], k);
    else return kth(ch[x][1], k-r);
}
void split(int l, int r){
    int x = kth(rt, l), y = kth(rt, r+2);
    splay(x, rt); splay(y, ch[rt][1]);
}
void rever(int l, int r){
    split(l, r);
    rev[ch[ch[rt][1]][0]] ^= 1;
}
void add(int l, int r, int v){
    split(l, r);
    tag[ch[ch[rt][1]][0]] += v;
    val[ch[ch[rt][1]][0]] += v;
    push_up(ch[ch[rt][1]][0]);
}
int build(int l, int r, int f){
    if(l > r) return 0;
    if(l == r){
        fa[l] = f;
        sz[l] = 1;
        return l;
    }
    int mid = l + r >> 1;
    ch[mid][0] = build(l, mid-1, mid);
    ch[mid][1] = build(mid+1, r, mid);
    fa[mid] = f;
    push_up(mid);
    return mid;
}
int asksum(int l, int r){
    split(l, r);
    return sum[ch[ch[rt][1]][0]];
}
int main(){
    //總共兩個數
    n = 2;
    rt = build(1, n+2, 0);//建樹
    for(int i = 1; i <= n; i++){
        scanf("%d", &x);
        add(i, i, x);//區間加
    }
    rever(1, n);//區間翻轉
    printf("%d\n", asksum(1, n));//區間求和
    return 0;
}

14.spfa

#include<cstdio>
using namespace std;
int n,m,a,b,op,head[200009],next[200009],dis[200009],len[200009],v[200009],l,r,team[200009],pd[100009],u,v1,e;
int lt(int x,int y,int z)
{
    op++,v[op]=y;
    next[op]=head[x],head[x]=op,len[op]=z;
}
int SPFA(int s,int f)//SPFA……
{
    for(int i=1;i<=200009;i++){dis[i]=999999999;}
    l=0,r=1,team[1]=s,pd[s]=1,dis[s]=0;
    while(l!=r)
    {
        l=(l+1)%90000,u=team[l],pd[u]=0,e=head[u];
        while(e!=0)
        {
            v1=v[e];
            if(dis[v1]>dis[u]+len[e])
            {
                dis[v1]=dis[u]+len[e];
                if(!pd[v1])
                {
                    r=(r+1)%90000,
                    team[r]=v1,
                    pd[v1]=1;
                }
            }
            e=next[e];
        } 
    }
    return dis[f];
}
int main()
{
    scanf("%d%d",&a,&b);
    lt(1,2,a);lt(2,3,b);//1到2為a，2到3為b，1到3即為a+b……
    printf("%d",SPFA(1,3));
    return 0;
}

15.floyd

#include<iostream>
#include<cstring>
using namespace std;
long long n=3,a,b,dis[4][4];
int main()
{
    cin>>a>>b;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            dis[i][j]=2147483647;
        }
    }
    dis[1][2]=a,dis[2][3]=b;
    for(int k=1;k<=n;k++)
    {
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);//Floyd……
            }
        }
    }
    cout<<dis[1][3];
}

16.Dijkstra+STL的優先佇列優化

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <climits>
#include <algorithm>
#include <map>
#include <queue>
#include <vector>
#include <ctime>
#include <string>
#include <cstring>
using namespace std;
const int N=405;
struct Edge {
    int v,w;
};
vector<Edge> edge[N*N];
int n;
int dis[N*N];
bool vis[N*N];
struct cmp {
    bool operator()(int a,int b) {
        return dis[a]>dis[b];
    }
};
int Dijkstra(int start,int end)
{
    priority_queue<int,vector<int>,cmp> dijQue;
    memset(dis,-1,sizeof(dis));
    memset(vis,0,sizeof(vis));
    dijQue.push(start);
    dis[start]=0;
    while(!dijQue.empty()) {
        int u=dijQue.top();
        dijQue.pop();
        vis[u]=0;
        if(u==end)
            break;
        for(int i=0; i<edge[u].size(); i++) {
            int v=edge[u][i].v;
            if(dis[v]==-1 || dis[v]>dis[u]+edge[u][i].w) {
                dis[v]=dis[u]+edge[u][i].w;
                if(!vis[v]) {
                    vis[v]=true;
                    dijQue.push(v);
                }
            }
        }
    }
    return dis[end];
}
int main()
{
    int a,b;
    scanf("%d%d",&a,&b);
    Edge Qpush;

    Qpush.v=1;
    Qpush.w=a;
    edge[0].push_back(Qpush);

    Qpush.v=2;
    Qpush.w=b;
    edge[1].push_back(Qpush);

    printf("%d",Dijkstra(0,2));
    return 0;
}

17.LCA

#include<cstdio>                                                  //標頭檔案
#define NI 2                                                          
//從來不喜歡算log所以一般用常數 不知道算不算壞習慣 因為3個節點 所以log3(當然以2為底)上取整得2
struct edge
{
    int to,next,data;                                              //分別表示邊的終點，下一條邊的編號和邊的權值
}e[30];                                                                     //鄰接表，點少邊少開30是為了浪啊
int v[10],d[10],lca[10][NI+1],f[10][NI+1],tot=0;      //陣列開到10依然為了浪
//陣列還解釋嘛，v表示第一條邊在鄰接表中的編號，d是深度，lca[x][i]表示x向上跳2^i的節點，f[x][i]表示x向上跳2^i的距離和
void build(int x,int y,int z)                                      //建邊
{
    e[++tot].to=y; e[tot].data=z; e[tot].next=v[x]; v[x]=tot;
    e[++tot].to=x; e[tot].data=z; e[tot].next=v[y]; v[y]=tot;
}
void dfs(int x)                                                        //遞迴建樹
{
    for(int i=1;i<=NI;i++)                                   //懶，所以常數懶得優化
        f[x][i]=f[x][i-1]+f[lca[x][i-1]][i-1],
        lca[x][i]=lca[lca[x][i-1]][i-1];                   //建樹的同時進行預處理
    for(int i=v[x];i;i=e[i].next)                              //遍歷每個連線的點
    {
        int y=e[i].to;
        if(lca[x][0]==y) continue;
        lca[y][0]=x;                                       //小技巧：lca[x][0]即為x的父親~~（向上跳2^0=1不就是父節點嘛）
        f[y][0]=e[i].data;
        d[y]=d[x]+1;
        dfs(y);                                            //再以這個節點為根建子樹【這裡真的用得到嘛？？】
    }
}
int ask(int x,int y)                                             //詢問，也是關鍵
{                                                                        
    if(d[x]<d[y]) {int t=x;x=y;y=t;}                  //把x搞成深的點
    int k=d[x]-d[y],ans=0;
    for(int i=0;i<=NI;i++)
        if(k&(1<<i))                                      //若能跳就把x跳一跳
            ans+=f[x][i],                              //更新資訊
            x=lca[x][i];
    for(int i=NI;i>=0;i--)                                  //不知道能不能正著迴圈，好像倒著優，反正記得倒著就好了
        if(lca[x][i]!=lca[y][i])                            //如果x跳2^i和y跳2^j沒跳到一起就讓他們跳
            ans+=f[x][i]+f[y][i],
            x=lca[x][i],y=lca[y][i];
    return ans+f[x][0]+f[y][0];                           //跳到LCA上去（每步跳的時候都要更新資訊，而且要在跳之前更新資訊哦~）
}
int main()
{
    int a,b;
    scanf("%d%d",&a,&b);
    build(1,2,a);
    build(1,3,b);                                                       //分別建1 2、1 3之間的邊
    dfs(1);                                                                //以1為根建樹
    printf("%d",ask(2,3));                                         //求解2 3到它們的LCA的距離和並輸出
}

18.LCT

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstring>
using namespace std;
struct node 
{
    int data,rev,sum;
    node *son[2],*pre;
    bool judge();
    bool isroot();
    void pushdown();
    void update();
    void setson(node *child,int lr);
}lct[233];
int top,a,b;
node *getnew(int x)
{
    node *now=lct+ ++top;
    now->data=x;
    now->pre=now->son[1]=now->son[0]=lct;
    now->sum=0;
    now->rev=0;
    return now;
}
bool node::judge(){return pre->son[1]==this;}
bool node::isroot()
{
    if(pre==lct)return true;
    return !(pre->son[1]==this||pre->son[0]==this);
}
void node::pushdown()
{
    if(this==lct||!rev)return;
    swap(son[0],son[1]);
    son[0]->rev^=1;
    son[1]->rev^=1;
    rev=0;
}
void node::update(){sum=son[1]->sum+son[0]->sum+data;}
void node::setson(node *child,int lr)
{
    this->pushdown();
    child->pre=this;
    son[lr]=child;
    this->update();
}
void rotate(node *now)
{
    node *father=now->pre,*grandfa=father->pre;
    if(!father->isroot()) grandfa->pushdown();
    father->pushdown();now->pushdown();
    int lr=now->judge();
    father->setson(now->son[lr^1],lr);
    if(father->isroot()) now->pre=grandfa;
    else grandfa->setson(now,father->judge());
    now->setson(father,lr^1);
    father->update();now->update();
    if(grandfa!=lct) grandfa->update();
}
void splay(node *now)
{
    if(now->isroot())return;
    for(;!now->isroot();rotate(now))
    if(!now->pre->isroot())
    now->judge()==now->pre->judge()?rotate(now->pre):rotate(now);
}
node *access(node *now)
{
    node *last=lct;
    for(;now!=lct;last=now,now=now->pre)
    {
        splay(now);
        now->setson(last,1);
    }
    return last;
}
void changeroot(node *now)
{
    access(now)->rev^=1;
    splay(now);
}
void connect(node *x,node *y)
{
    changeroot(x);
    x->pre=y;
    access(x);
}
void cut(node *x,node *y)
{
    changeroot(x);
    access(y);
    splay(x);
    x->pushdown();
    x->son[1]=y->pre=lct;
    x->update();
}
int query(node *x,node *y)
{
    changeroot(x);
    node *now=access(y);
    return now->sum;
}
int main()
{
    scanf("%d%d",&a,&b);
    node *A=getnew(a);
    node *B=getnew(b);
    //連邊 Link
        connect(A,B);
    //斷邊 Cut
        cut(A,B);
    //再連邊orz Link again
        connect(A,B);
    printf("%d\n",query(A,B)); 
    return 0;
}

19.等差數列

#include<iostream>
using namespace std;
int s1,s2,s3,s4;
int n,m;
int main()
{
    cin>>n>>m;
    s1=(1+n)*n/2;    算出1+2+3+...+n;
    s2=(1+n-1)*(n-1)/2;  算出1+2+3+...+（n-1）;
    s3=(1+m)*m/2;    算出1+2+3+...+m；
    s4=(1+m-1)*(m-1)/2;   算出1+2+3+...+（m-1）；
    cout<<(s1-s2)+(s3-s4);   輸出就可以了。
    return 0;
}

20.高精-模板類

include <bits/stdc++.h>

using namespace std;

class cint{ //定義一個類

private:

int c_number[100001],c_len,c_d,c_fh; //屬性,包括數字,長度,進位制,符號

public:

        cint(); 
        ~cint();
        cint(int x); 
        cint(string st); //構造與解構函式
        cint operator+(cint& b); //過載+,以用於更方便地運算
        cint read_cint(); //讀入
        void write_cint(); //輸出
};
cint::cint()
{
    c_d=10;
}
cint::~cint()
{
}
cint::cint(int x)
{
    c_d=10;
    if (x<0)
    {
        c_fh=-1;
        x=-x;
    }
    else c_fh=1;
    c_len=0;
    while (x)
    {
        c_len++;
        c_number[c_len]=x%c_d;
        x/=c_d;
    }
}
cint::cint (string st)
{
    int i;
    if (st[0]=='-') 
    {
        c_fh=-1;
        st.erase(0,1);
    }
    else c_fh=1;
    while (st[0]=='0'&&st.length()>1)
        st.erase(0,1); //去除前導0
    c_len=st.length();
    for (i=1;i<=c_len;i++)
        c_number[i]=(st[c_len-i]-48)*c_fh; //將字元的ascii碼-48,存入陣列中
} //建構函式,將字串存入類中
cint cint::operator+(cint& b)
{
    int i;
    cint c;
    if (c_len>=b.c_len) c.c_len=c_len;
    else c.c_len=b.c_len;
    for (i=1;i<=c.c_len;i++)
    {
        c.c_number[i]+=c_number[i]+b.c_number[i]; //將兩位相加
        c.c_number[i+1]=c.c_number[i]/c.c_d;
        c.c_number[i]%=c.c_d; //處理進位
    }
    while (c.c_number[c.c_len+1])
        c.c_len++;
    return c;
} //核心部分,高精加
cint cint::read_cint()
{
    string st;
    cin>>st;
    return cint(st);
}
void cint::write_cint()
{
    int i;
    for (i=1;i<=c_len;i++)
        cout<<c_number[c_len-i+1]; //輸出部分,很容易理解
}
istream& operator>>(istream& is,cint &c)
{
    c=c.read_cint();
    return is;
} //過載>>,便於輸入
ostream& operator<<(ostream& os,cint c)
{
    c.write_cint();
    return os;
} //過載<<,便於輸出.
int main()
{
    cint a,b;
    cin>>a>>b;
    cout<<a+b<<endl;
}

21.fread & fwrite優化

#include <cstdio>
const size_t fSize = 1 << 15;
char iFile[fSize], *iP = iFile, oFile[fSize], *oP = oFile;
inline char readchar() {
    if (*iP && iP - iFile < fSize) { char t = *iP; iP++; return t; } else return EOF;
}
template<typename T> inline void readint(T &x) {
    x = 0; char c; bool neg = 0;
    while ((c = readchar()) < '0' || c > '9') if (c == '-') neg = !neg;
    while (c >= '0' && c <= '9')
        x = x * 10 + (c ^ 48), c = readchar();
    x = neg ? -x : x;
}
inline void writechar(const char &c) { *oP = c, ++oP; }
template<typename T> inline void _writeint(const T &x) {
    if (!x)
        return;
    _writeint(x / 10);
    writechar(x % 10 ^ 48);
}
template<typename T> inline void writeint(T x, const char &c) {
    if (x < 0) {
        writechar('-');
        x = -x;
    }
    if (!x) {
        writechar('0');
        return;
    }
    _writeint(x);
    writechar(c);
}
int main() {
    fread(iFile, 1, fSize, stdin);
    int a, b;
    readint(a); readint(b);
    writeint(a + b, '\n');
    fwrite(oFile, 1, oP - oFile, stdout);
    return 0;
}

22.位運算-非遞迴

#include <cstdio>
int m, n;
int main()
{
    scanf("%d%d", &m, &n);
    int u = m & n;
    int v = m ^ n;
    while (u) {
        int s = v;
        int t = u << 1;
        u = s & t;
        v = s ^ t;
    }
    printf("%d\n", v);
}

23.模擬

#include <iostream> 
#include <cmath>
using namespace std;
int fu=1,f=1,a,b,c=0;
int main()
{
    cin>>a>>b;
    if(a<0&&b>0)fu=2;
    if(a>0&&b<0)fu=3;
    if(a<0&&b<0)f=-1;
    if(a==0){cout<<b;return 0;}
    if(b==0){cout<<a;return 0;} 
    a=abs(a);
    b=abs(b);
    if(a>b&&fu==3)f=1;
    if(b>a&&fu==3)f=-1;
    if(b>a&&fu==2)f=1;
    if(b<a&&fu==2)f=-1;
    if(fu==1)c=a+b;
    if(fu>1)c=max(a,b)-min(a,b);
    c*=f;
    cout<<c;
    return 0;
}