連結：http://codeforces.com/contest/1082

A. Vasya and Book

題意：

n,x,y,d
一本電子書有n頁，每一次翻動只能往前或者往後翻d頁。求x->y頁最少需要多少步。只能在（1~n)之間翻。具體細節看題目吧。部落格僅作記錄

int t,n,x,y,d;
int calc(int x,int y)
{
    return abs(x-y)/d;
}
int main() 
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d%d",&n,&x,&y,&d);
        int res = abs(x-y);
        if(res%d==0)
        {
            cout<<res/d<<endl;continue;
        }
        int ans = inf;
        if((y-1)%d==0)
        {
            ans = min(ans,(int)ceil((x-1.)/d)+(y-1)/d);
        }
        if((n-y)%d==0)
        {
            ans = min(ans,(int)ceil((double)(n-x)/d)+(n-y)/d);
        }
        if(ans == inf)
            ans = -1;
        cout<<ans<<endl;
    }
    return 0;
}
 

B. Vova and Trophies

渣渣的程式碼：

char s[200010];
int d[200010];
int n;
int main() 
{
    scanf("%d",&n);
    scanf("%s",s+1);
    int num = 0;
    for(int i=1;i<=n;i++)
        if(s[i]=='G')
            num++;
    int ans = 0;
    for(int i=1;i<=n;i++)
    {
        if(s[i]!='G')
        {
            d[i] = 0;
        }
        else
        {
            d[i] = 1;
            if(s[i-1] == 'G')
                d[i] = d[i-1]+1;
        }
    }
    for(int i=1;i<=n;i++)
    {
        ans = max(ans,d[i]);
        if(s[i-d[i]]=='S'&&d[i]<num)
        {
            if(s[i-d[i]-1]=='G')
            {
                if(d[i]+d[i-d[i]-1]<num)
                ans = max(ans,d[i]+1+d[i-d[i]-1]);
                else
                    ans = max(ans,d[i]+d[i-d[i]-1]);
            }
            ans = max(ans,d[i]+1);
        }
    }
    printf("%d\n",ans);
    return 0;
}
 

大神的程式碼

#include <bits/stdc++.h>
using namespace std;
int res,pre,cnt,g;
int main()
{
    int n;
    cin>>n;
    while(n--)
    {
        char s;
        cin>>s;
        if(s=='G')cnt++,g++;
        else pre = cnt,cnt = 0;
        res = max(res,cnt+pre+1);
    }
    cout<<min(g,res);
    return 0;
}
 

C. Multi-Subject Competition

vector<int> s[100010];
int n,m;

bool cmp(int a,int b)
{
    return a>b;
}
bool cmp2(vector<int>a,vector<int> b)
{
    return a.size()>b.size();
}
int main() 
{
    cin>>n>>m;
    int t,r;
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d",&t,&r);
        s[t].push_back(r);
    }
    int mi = 0;
    for(int i=1;i<=m;i++)
    {
        sort(s[i].begin(),s[i].end(),cmp);
        for(int j=1;j<s[i].size();j++)
            s[i][j] += s[i][j-1];
        mi = max(mi,(int)s[i].size());
    }
    sort(s+1,s+m+1,cmp2);
    int ans = 0;
    for(int i=0;i<mi;i++)
    {
        int sum = 0;
        for(int j=1;j<=m;j++)
        {
            if(i>=s[j].size())break;
            sum = max(sum,sum+s[j][i]);
            //printf("%d %d\n",j,s[j][i]);
        }
        //cout<<sum<<endl;
        ans = max(ans,sum);
    }
    cout<<ans<<endl;
    return 0;
}

大神的程式碼

#include<bits/stdc++.h>
using namespace std;
long long n,m,s,r,k,c,mx,a[200000],l;
pair<long long,long long> p[200000];
int main()
{
    cin>>n>>m;
    for(int i=0;i<n;i++)
    {
        cin>>s>>r;
        p[i]={s,-r};
    }
    sort(p,p+n);
    for(int i=0;i<n;i++)
    {
        if(p[i].first!=l)
        {
            k=0; c=0; l=p[i].first;
        }
        c-=p[i].second;
        k++;
        if(c>0)
            a[k]+=c;
        mx=max(mx,a[k]);
    }
    cout<<mx;
    return 0;
}