Codeforces

Number of Ways

题目链接

分类：思维、暴力

1.题意概述

• 给你n个数a[1…n]，要你找符合条件的i,j(2≤i≤j≤n−1)，满足∑k=1i−1ak=∑k=ijak=∑k=j+1nak，求符合条件的i,j的组合数？

2.解题思路

• 本意就是要你找三段使得他们和相等，那么数列首先得能被分成三段，也就是sum%3=0才行，然后再按组合统计即可，细节可以参见代码。

3.AC代码

#include <bits/stdc++.h>#define INF 0x3f3f3f3f#define maxn 500010#define lson root << 1#define rson root << 1 | 1#define lent (t[root].r - t[root].l + 1)#define lenl (t[lson].r - t[lson].l + 1)#define lenr (t[rson].r - t[rson].l + 1)#define N 1111#define eps 1e-6#define pi acos(-1.0)#define e exp(1.0)using namespace std;const int mod = 1e9 + 7;typedef long long ll;typedef unsigned long long ull;int a[maxn];ll sum[maxn];int main(){#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);    freopen("out.txt", "w", stdout);    long _begin_time = clock();#endif    int n;    scanf("%d", &n);    sum[0] = 0;    for (int i = 1; i <= n; i++)    {        scanf("%d", &a[i]);        sum[i] = sum[i - 1] + a[i];    }    ll cnt = 0;    if (sum[n] % 3 == 0)    {        ll tmp1 = sum[n] / 3;        ll tmp2 = tmp1 << 1;        int count = 0;        for (int i = 1; i < n; i++)        {               if (sum[i] == tmp2)                cnt += count;            if (sum[i] == tmp1)                count++;        }    }    printf("%I64d\n", cnt);#ifndef ONLINE_JUDGE    long _end_time = clock();    printf("time = %ld ms.", _end_time - _begin_time);#endif    return 0;}