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
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 ",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); }