zoukankan      html  css  js  c++  java
  • Code Forces 21C Stripe 2

    C. Stripe 2
    time limit per test
    1 second
    memory limit per test
    64 megabytes
    input
    standard input
    output
    standard output

    Once Bob took a paper stripe of n squares (the height of the stripe is 1 square). In each square he wrote an integer number, possibly negative. He became interested in how many ways exist to cut this stripe into three pieces so that the sum of numbers from each piece is equal to the sum of numbers from any other piece, and each piece contains positive integer amount of squares. Would you help Bob solve this problem?

    Input

    The first input line contains integer n (1 ≤ n ≤ 105) — amount of squares in the stripe. The second line contains n space-separated numbers — they are the numbers written in the squares of the stripe. These numbers are integer and do not exceed 10000 in absolute value.

    Output

    Output the amount of ways to cut the stripe into three non-empty pieces so that the sum of numbers from each piece is equal to the sum of numbers from any other piece. Don't forget that it's allowed to cut the stripe along the squares' borders only.

    Examples
    input
    4
    1 2 3 3
    
    output
    1
    
    input
    5
    1 2 3 4 5
    
    output
    0

    切割两个点,那么两个点到顶点的区间和肯定是总和的3分之1.对于每个点找到他右边有多少个满足条件的点,

    #include <iostream>
    #include <string.h>
    #include <stdlib.h>
    #include <algorithm>
    #include <math.h>
    #include <stdio.h>
    
    using namespace std;
    int n;
    long long int a[100005];
    long long int s[100005];
    int l[100005];
    int r[100005];
    long long int sum;
    int main()
    {
        scanf("%d",&n);
        s[0]=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&a[i]);
            s[i]=s[i-1]+a[i];
            sum+=a[i];
        }
        if(sum%3!=0||n<=2)
        {printf("0
    ");return 0;}
        int cnt=0;int cot=0;
        memset(l,0,sizeof(l));
        memset(r,0,sizeof(r));
        for(int i=n;i>=1;i--)
        {
            if(s[n]-s[i-1]==sum/3)
            {
                r[i]=r[i+1]+1;
            }
            else
                r[i]=r[i+1];
        }
        long long int ans=0;
        for(int i=1;i<=n;i++)
        {
            if(s[i]==sum/3)
            {
                int x=r[i+2];
    
                ans+=x;
            }
        }
        printf("%lld
    ",ans);
        return 0;
    
    }
    


  • 相关阅读:
    [HDU2866] Special Prime (数论,公式)
    [骗分大法好] 信息学竞赛 骗分导论(论文搬运)
    flayway数据库管理
    RabbitMQ的基本概念与原理
    springboot+ideal实现远程调试
    盘点总结
    mysql查看进程命令
    Java字符串正则文本替换
    springboot代码级全局敏感信息加解密和脱敏方案
    使用PMD进行代码审查
  • 原文地址:https://www.cnblogs.com/dacc123/p/8228718.html
Copyright © 2011-2022 走看看