P3131 [USACO16JAN]子共七Subsequences Summing to Sevens

题目描述

Farmer John's NN 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 \ldots 61…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.

给你n个数，求一个最长的区间，使得区间和能被7整除

输入输出格式

输入格式：

The first line of input contains NN (1 \leq N \leq 50,0001≤N≤50,000). The next NN

lines each contain the NN integer IDs of the cows (all are in the range

0 \ldots 1,000,0000…1,000,000).

输出格式：

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.

输入输出样例

输入样例#1：复制

输出样例#1：复制

说明

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

思路：前缀和+二分答案。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int n,l,r,mid;
int num[50010],sum[50010];
bool judge(){for(int i=0;i<=n-mid;i++)if((sum[i+mid]-sum[i])%7==0)    return true;return false;
}
int main(){scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&num[i]),sum[i]=sum[i-1]+num[i];l=1;r=n;while(l<=r){mid=(l+r)/2;if(judge())    l=mid+1;else r=mid-1;}cout<<l-1;
}

思路：求出前缀和mod7，然后遍历，如果拥有相同的余数，说明这个区间是可以被7整除的记录。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define MAXN 50010
using namespace std;
int n;
int pri[10],v[10];
int a[MAXN],sum[MAXN];
int main(){scanf("%d",&n);sum[0]=0;for(int i=1;i<=n;i++)scanf("%lld",&a[i]),sum[i]=(sum[i-1]+a[i])%7;for(int i=1;i<=n;i++){if(!v[sum[i]])v[sum[i]]=i,pri[sum[i]]=i;else    pri[sum[i]]=i;}int ans=-1;for(int i=0;i<7;i++){if(!v[i])    continue;ans=max(ans,pri[i]-v[i]);}printf("%d\n",ans);
}

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define MAXN 50010
using namespace std;
int n;
int pri[10],v[10];
int a[MAXN],sum[MAXN];
int main(){scanf("%d",&n);sum[0]=0;for(int i=1;i<=n;i++)scanf("%lld",&a[i]),sum[i]=(sum[i-1]+a[i])%7;for(int i=1;i<=n;i++){if(!v[sum[i]])v[sum[i]]=i,pri[sum[i]]=i;else    pri[sum[i]]=i;}int ans=-1;for(int i=0;i<7;i++){if(!v[i])    continue;ans=max(ans,pri[i]-v[i]);}printf("%d\n",ans);
}

转载于:https://www.cnblogs.com/cangT-Tlan/p/7862498.html

洛谷 P3131 [USACO16JAN]子共七Subsequences Summing to Sevens相关推荐

[USACO16JAN]子共七Subsequences Summing to Sevens
题目描述 Farmer John's NNN cows are standing in a row, as they have a tendency to do from time to time. ...
题161.洛谷P3131 前缀和与差分-Subsequences Summing to Sevens S
文章目录题161.洛谷P3131 前缀和与差分-Subsequences Summing to Sevens S 一.题目二.题解题161.洛谷P3131 前缀和与差分-Subsequences ...
洛谷P3131 [USACO16JAN]Subsequences Summing to Sevens S 题解
洛谷P3131 [USACO16JAN]Subsequences Summing to Sevens S 题解题目链接:P3131 [USACO16JAN]Subsequences Summing ...
洛谷P3131 [USACO16JAN]Subsequences Summing to Sevens S
题目链接:P3131 [USACO16JAN]Subsequences Summing to Sevens S - 洛谷 | 计算机科学教育新生态 (luogu.com.cn) 题目大意: 看到英文题 ...
一中OJ #3509 七的倍数 [USACO Jan16,洛谷P3131] | 同余前缀和 | 解题报告
一中OJ | #3509 七的倍数 [USACO Jan16 Silver , Subsequences Summing to Sevens] 时限 1000MS/Case 内存 128MB/Case ...
P3131 [USACO16JAN]Subsequences Summing to Sevens S
P3131 [USACO16JAN]Subsequences Summing to Sevens S 提交13.65k 通过3.63k 时间限制200ms 内存限制128.00MB 提交答案加入题单 ...
题目：P3131 [USACO16JAN]Subsequences Summing to Sevens S
题目:[USACO16JAN]Subsequences Summing to Sevens S - 洛谷题目大意给定一个序列,要求计算出能被7整除的最长序列坑点无数据范围 int足以思路 ...
[洛谷 P3788] 幽幽子吃西瓜
妖梦费了好大的劲为幽幽子准备了一个大西瓜,甚至和兔子铃仙打了一架.现在妖梦闲来无事,就蹲在一旁看幽幽子吃西瓜.西瓜可以看作一个标准的球体,瓜皮是绿色的,瓜瓤是红色的,瓜皮的厚度可视为0.妖梦恰好以正视 ...
P3131 [USACO16JAN]Subsequences Summing to Sevens S-二分+前缀和
给你n个数,分别是a[1],a[2],-,a[n].求一个最长的区间[x,y],使得区间中的数(a[x],a[x+1],a[x+2],-,a[y-1],a[y])的和能被7整除.输出区间长度.若没有符 ...

洛谷 P3131 [USACO16JAN]子共七Subsequences Summing to Sevens

P3131 [USACO16JAN]子共七Subsequences Summing to Sevens

题目描述

输入输出格式

输入输出样例

说明

洛谷 P3131 [USACO16JAN]子共七Subsequences Summing to Sevens相关推荐

最新文章

热门文章