1006 Magic Pen 6

題目的意思就是給你一串數字，讓你找出其中最長的一個串使得這些數字之和模M等於0：

那就直接先計算出sum(i) = a1+a2+....+ai;一串數字的和就是這串數字的首數字和末尾數字的sum之差,要使得字串

最長，那麼就使用兩個指標，一個從頭開始，一個從尾部開始，只要找到滿足條件的數字串就直接跳出，這時候的長度

肯定就是最長的了：

#include <iostream>
#include <cstdio>
#include <string>
#include <string.h>
#include <map>
#include <vector>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <queue>
#include <set>
#include <stack>
#include <functional>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cassert>
#include <bitset>
#include <stack>
#include <ctime>
#include <list>
#define INF 0x7fffffff
#define max3(a,b,c) (max(a,b)>c?max(a,b):c)
#define min3(a,b,c) (min(a,b)<c?min(a,b):c)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;

#define N 100005
int a[N] ;
__int64 sum[N] ;
int main()
{
    int n , m ;
    while(cin >> n >> m)
    {
        mem(sum ,0) ;
        for (int i = 1 ; i <= n ; i ++ )
        {
            scanf("%d",&a[i]) ;
            sum[i] = sum[i - 1] + a[i] ;
        }
        __int64 l = 1 , r = n ;
        __int64 ed ;
        __int64 ans = 0 ;
        __int64 c = sum[n] % m ;
        if(sum[n] % m == 0)
        {
            cout << n << endl;
            continue ;
        }

        int flag = 0 ;
        for (int ed = n ; ed >= 1 ; --ed )
        {
            for (int st = 1 ; st + ed - 1 <= n ; ++st)
            {
                if((sum[st + ed - 1] - sum[st - 1]) % m == 0)
                {
                    cout << ed << endl;
                    flag = 1 ;
                    break ;
                }
            }
            if(flag)
                break ;
        }
        if(!flag)
            cout << 0 << endl;
    }
    return 0 ;
}
 

1005 Another Graph Game

這道題目一開始想複雜了，想到了刪邊遊戲，太複雜了。既然就是求差，那麼可以分情況討論一下：

若沒有邊權，則對點權從大到小排序即可。

考慮邊，將邊權拆成兩半加到它所關聯的兩個點的點權中即可。因為當兩個人分別選擇不同的點時，這一權值將互相抵消。

#include <iostream>
#include <cstdio>
#include <string>
#include <string.h>
#include <map>
#include <vector>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <queue>
#include <set>
#include <stack>
#include <functional>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cassert>
#include <bitset>
#include <stack>
#include <ctime>
#include <list>
#define INF 0x7fffffff
#define max3(a,b,c) (max(a,b)>c?max(a,b):c)
#define min3(a,b,c) (min(a,b)<c?min(a,b):c)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
#define N 100001
int n, m;
double num[N];

int a, b;
double w;
bool cpp(double a, double b)
{
    return a > b;
}
double ans = 0;
int main()
{
    while(scanf("%d%d", &n, &m) == 2)
    {
        mem(num, 0);
        ans = 0;
        for(int i = 0; i < n; ++i)
            scanf("%lf", &num[i]);
        for(int i = 0; i < m; ++i)
        {
            scanf("%d%d%lf", &a, &b, &w);
            num[a-1] += w/2.0;
            num[b-1] += w/2.0;
        }
        sort(num, num + n, cpp);

        double la = 0, lb = 0;
        for(int i = 0; i < n ; i = i + 2)
        {
            ans += num[i] - num[i+1];
        }
        printf("%.0lf\n",ans);
    }

    return 0;
}
 

數論？ 真的不會

題目大意：將一個數n拆分成幾個數相加，問有多少種方案？

思路：

#include <iostream>
#include <cstdio>
#include <string>
#include <string.h>
#include <map>
#include <vector>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <queue>
#include <set>
#include <stack>
#include <functional>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cassert>
#include <bitset>
#include <stack>
#include <ctime>
#include <list>
#define INF 0x7fffffff
#define max3(a,b,c) (max(a,b)>c?max(a,b):c)
#define min3(a,b,c) (min(a,b)<c?min(a,b):c)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;

const int mod = 1e9+7;
#define N 100020

int p[N];
void Partition()
{
    p[0] =1;
    for(int n = 1; n <= 1e5; ++n)
    {
        int fac = 1;
        int  k =1;
        while(1)
        {
            int t = n - k * (3 * k - 1) / 2;
            if(t < 0)
                break;
            p[n] = (p[n] + fac * p[t])%mod;
            if(t-k >= 0)
                p[n] = (p[n] + fac*p[t-k])%mod;
            p[n] %= mod;
            fac = -fac;
            ++k;
        }
        p[n] = (p[n] + mod) % mod;
    }
}
int main()
{
    Partition();
    int d;
    scanf("%d",&d);
    while(d--)
    {
        int c;
        scanf("%d",&c);
        printf("%d\n",p[c]);
    }
    return 0;
}