题目链接:
sum
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 131072/131072 K (Java/Others)
Problem Description
Given a sequence, you're asked whether there exists a consecutive subsequence whose sum is divisible by m. output YES, otherwise output NO
Input
The first line of the input has an integer T (1≤T≤10), which represents the number of test cases.
For each test case, there are two lines:
1.The first line contains two positive integers n, m (1≤n≤100000, 1≤m≤5000).
2.The second line contains n positive integers x (1≤x≤100) according to the sequence.
For each test case, there are two lines:
1.The first line contains two positive integers n, m (1≤n≤100000, 1≤m≤5000).
2.The second line contains n positive integers x (1≤x≤100) according to the sequence.
Output
Output T lines, each line print a YES or NO.
Sample Input
2
3 3
1 2 3
5 7
6 6 6 6 6
Sample Output
YES
NO
题意:
问是否存在字串的和为m倍数;
思路:
sum[r]-sum[l]=k*m;sum[r]=sum[l](modm);
前缀和同余就存在;
AC代码:
/************************************************ ┆ ┏┓ ┏┓ ┆ ┆┏┛┻━━━┛┻┓ ┆ ┆┃ ┃ ┆ ┆┃ ━ ┃ ┆ ┆┃ ┳┛ ┗┳ ┃ ┆ ┆┃ ┃ ┆ ┆┃ ┻ ┃ ┆ ┆┗━┓ ┏━┛ ┆ ┆ ┃ ┃ ┆ ┆ ┃ ┗━━━┓ ┆ ┆ ┃ AC代马 ┣┓┆ ┆ ┃ ┏┛┆ ┆ ┗┓┓┏━┳┓┏┛ ┆ ┆ ┃┫┫ ┃┫┫ ┆ ┆ ┗┻┛ ┗┻┛ ┆ ************************************************ */ #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <bits/stdc++.h> #include <stack> using namespace std; #define For(i,j,n) for(int i=j;i<=n;i++) #define mst(ss,b) memset(ss,b,sizeof(ss)); typedef long long LL; template<class T> void read(T&num) { char CH; bool F=false; for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar()); for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar()); F && (num=-num); } int stk[70], tp; template<class T> inline void print(T p) { if(!p) { puts("0"); return; } while(p) stk[++ tp] = p%10, p/=10; while(tp) putchar(stk[tp--] + '0'); putchar(' '); } const LL mod=1e9+7; const double PI=acos(-1.0); const int inf=1e9; const int N=1e5+10; const int maxn=(1<<8); const double eps=1e-8; int vis[5600]; int main() { int t; read(t); while(t--) { mst(vis,0); int n,m,sum=0,flag=0,x; read(n);read(m); vis[0]=1; For(i,1,n) { read(x); sum+=x; int temp=sum%m; if(vis[temp])flag=1; vis[temp]=1; } if(flag)cout<<"YES "; else cout<<"NO "; } return 0; }