codeforces377B Modulo Sum(抽屉+dp)
2015-10-22 22:30
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You are given a sequence of numbers a1, a2, ..., an, and a number m.
Check if it is possible to choose a non-empty subsequence aij such that the sum of numbers in this subsequence is divisible by m.
Input
The first line contains two numbers, n and m (1 ≤ n ≤ 106, 2 ≤ m ≤ 103) — the size of the original sequence and the number such that sum should be divisible by it.
The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
Output
In the single line print either "YES" (without the quotes) if there exists the sought subsequence, or "NO" (without the quotes), if such subsequence doesn't exist.
Sample test(s)
input
3 5
1 2 3
output
YES
input
1 6
5
output
NO
input
4 6
3 1 1 3
output
YES
input
6 6
5 5 5 5 5 5
output
YES
Note
In the first sample test you can choose numbers 2 and 3, the sum of which is divisible by 5.
In the second sample test the single non-empty subsequence of numbers is a single number 5. Number 5 is not divisible by 6, that is, the sought subsequence doesn't exist.
In the third sample test you need to choose two numbers 3 on the ends.
In the fourth sample test you can take the whole subsequence.
更新----我是傻逼,我是傻逼,我是傻逼,看一个知识点总是粗略的看一下,而后的原理自己都搞的很少。
鸽巢就是告诉了我们n>=m时是一定有的,而我呢?更新看代码吧,,,看过别人的代码
---------完结
也许是自己菜吧,反正觉得这是一道好题。
首先当n>=m时我们直接套用抽屉原理判断,对于n<m的现象自己最先没有想到,看了别人的题解才明白了需要用dp。
dp[i][j]记录的是i个数里,和为j的数字有没有出现过,最后判断dp
[c]是否等于1
这是自己直接套用的抽屉原理模板,应该可以改简单的,明天再看一下。
更新:
Check if it is possible to choose a non-empty subsequence aij such that the sum of numbers in this subsequence is divisible by m.
Input
The first line contains two numbers, n and m (1 ≤ n ≤ 106, 2 ≤ m ≤ 103) — the size of the original sequence and the number such that sum should be divisible by it.
The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
Output
In the single line print either "YES" (without the quotes) if there exists the sought subsequence, or "NO" (without the quotes), if such subsequence doesn't exist.
Sample test(s)
input
3 5
1 2 3
output
YES
input
1 6
5
output
NO
input
4 6
3 1 1 3
output
YES
input
6 6
5 5 5 5 5 5
output
YES
Note
In the first sample test you can choose numbers 2 and 3, the sum of which is divisible by 5.
In the second sample test the single non-empty subsequence of numbers is a single number 5. Number 5 is not divisible by 6, that is, the sought subsequence doesn't exist.
In the third sample test you need to choose two numbers 3 on the ends.
In the fourth sample test you can take the whole subsequence.
更新----我是傻逼,我是傻逼,我是傻逼,看一个知识点总是粗略的看一下,而后的原理自己都搞的很少。
鸽巢就是告诉了我们n>=m时是一定有的,而我呢?更新看代码吧,,,看过别人的代码
---------完结
也许是自己菜吧,反正觉得这是一道好题。
首先当n>=m时我们直接套用抽屉原理判断,对于n<m的现象自己最先没有想到,看了别人的题解才明白了需要用dp。
dp[i][j]记录的是i个数里,和为j的数字有没有出现过,最后判断dp
[c]是否等于1
这是自己直接套用的抽屉原理模板,应该可以改简单的,明天再看一下。
/* *********************************************** Author :Mosu Created Time :2015/10/10 18:30:45 File Name :\Users\Mosu\Desktop\project\BC.cpp ************************************************ */ #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <time.h> using namespace std; #define INF 0x7ffffff int neigb[1000009]; int dp[1005][1005]; long long S; struct Remnant{ int h,r; }R[1000009]; bool cmp(const Remnant &a,const Remnant &b){ if(a.r==b.r) return a.h<b.h; return a.r<b.r; } int main(){ int c,n,k,h; while(scanf("%d%d",&n,&c)!=EOF){ for(int i=0;i<n;i++) scanf("%d",&neigb[i]); if(n>c){ k=-1; S=0; for(int i=0;i<n;i++){ //scanf("%d",&neigb[i]); S+=neigb[i]; R[i].r=S%c; R[i].h=i+1; if(k==-1&&R[i].r==0) k=i; } if(k==-1){ sort(R,R+n,cmp); for(int i=0;i<n-1;i++){ if(k==-1&&R[i].r==R[i+1].r){ k=R[i].h; h=R[i+1].h; } } if(k==-1) printf("NO\n"); else{ printf("YES\n"); } } else{ printf("YES\n"); } } else{ for(int i=0;i<=n;i++) dp[i][0]=1; for(int i=1;i<=n;i++) for(int j=1;j<=c;j++) dp[i][j]=(dp[i-1][(j+neigb[i-1])%c]||dp[i-1][j]); if(dp [c]) cout<<"YES"<<endl; else cout<<"NO"<<endl; } } return 0; }
更新:
/* *********************************************** Author :Mosu Created Time :2015/10/10 18:30:45 File Name :\Users\Mosu\Desktop\project\BC.cpp ************************************************ */ #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <time.h> using namespace std; #define INF 0x7ffffff int dp[1009][1009]; int s[1000009]; int main() { int n,m; while(scanf("%d%d",&n,&m)==2){ int k=1; for(int i=0;i<n;i++){ scanf("%d",&s[i]); if(s[i]%m==0) k=0; } if(n>m||k==0){ printf("YES\n"); continue; } sizeof(dp,0,sizeof(dp)); for(int i=0;i<=n;i++) dp[i][0]=1; for(int i=1;i<=n;i++) for(int j=0;j<=m;j++) dp[i][j]=(dp[i-1][(j+s[i-1])%m]||dp[i-1][j]); if(dp [m]) printf("YES\n"); else printf("NO\n"); } return 0; }
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