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  • 洛谷 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 616, 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 50,0001N50,000). The next NN

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

    0 ldots 1,000,00001,000,000).

     

    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
    7
    3
    5
    1
    6
    2
    14
    10
    输出样例#1: 复制
    5

    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;
    }
    80

    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
    ",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
    ",ans);
    }
    细雨斜风作晓寒。淡烟疏柳媚晴滩。入淮清洛渐漫漫。 雪沫乳花浮午盏,蓼茸蒿笋试春盘。人间有味是清欢。
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  • 原文地址:https://www.cnblogs.com/cangT-Tlan/p/7862498.html
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