Subsequences Summing to Sevens(前缀和)

最新推荐文章于 2025-03-23 10:36:40 发布

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最新推荐文章于 2025-03-23 10:36:40 发布

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分类专栏：简单题文章标签：前缀和

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本文介绍了一种算法问题，即寻找整数序列中满足特定条件（元素之和为7的倍数）的最大连续子序列，并提供了完整的代码实现及解析。

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Subsequences Summing to Sevens

时间限制: 1 Sec 内存限制: 64 MB
提交: 55 解决: 29
[ 提交][ 状态][ 讨论版]

题目描述

Farmer John's N cows are standing in a row, as they have a tendency to do from time to time. Each cow is labeled with a distinct integer ID number so FJ can tell them apart. FJ would like to take a photo of a contiguous group of cows but, due to a traumatic childhood incident involving the numbers 1…6, he only wants to take a picture of a group of cows if their IDs add up to a multiple of 7.

Please help FJ determine the size of the largest group he can photograph.

输入

The first line of input contains N (1≤N≤50,000). The next N lines each contain the N integer IDs of the cows (all are in the range 0…1,000,000).

输出

Please output the number of cows in the largest consecutive group whose IDs sum to a multiple of 7. If no such group exists, output 0.

You may want to note that the sum of the IDs of a large group of cows might be too large to fit into a standard 32-bit integer. If you are summing up large groups of IDs, you may therefore want to use a larger integer data type, like a 64-bit "long long" in C/C++.

样例输入

样例输出

提示

In this example, 5+1+6+2+14 = 28.

又是一道前缀和的问题，没有修改插入，只是静态的查询，用前缀和进行预处理简直给力。

通过预处理前缀和mod7，放到sum数组中，如果两个位置的sum相等，即他们取余7之后的结果相同，则他们的差，也就是后面减去前面，剩下的部分，取余7等于0。

遍历取最长的就好了。

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#define pi acos(-1.0)
#define maxn (101000 + 50)
#define mol 1000000009
#define inf 0x3f3f3f3f
#define Lowbit(x) (x & (-x))
using namespace std;
typedef long long int LLI;

int a[maxn],sum[maxn];
int vis[10],n,re;

int main() {
   memset(vis,-1,sizeof(vis));
   int n;
   while(scanf("%d",&n)!=EOF){
        int tmp;
        for(int i=1;i<=n;i++){
            scanf("%d",&tmp);
            sum[i] = (sum[i-1] + tmp)% 7;
        }
        int re = 0;
        for(int i=1;i<=n;i++){
            if(vis[sum[i]] == -1){
                vis[sum[i]] = i;
            }
            else re = max(re , i - vis[sum[i]]);
        }
        printf("%d\n",re);
   }
}