codeforces377B Modulo Sum（抽屉+dp）

最新推荐文章于 2019-06-09 14:50:00 发布

Mosu_

最新推荐文章于 2019-06-09 14:50:00 发布

阅读量844

点赞数

分类专栏： Codeforces

本文链接：https://blog.youkuaiyun.com/Mosu_/article/details/49339915

版权

Codeforces 专栏收录该内容

9 篇文章

订阅专栏

博主反思自己在解决Codeforces 377B问题时，初看忽视了某些关键点。当n>=m时，利用抽屉原理直接得出结论。对于n<m的情况，通过学习他人题解，理解到需使用动态规划（DP）求解。博主分享了DP[i][j]的状态转移思路，并表示会进一步简化解题模板。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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[n][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[n][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[n][m])
			printf("YES\n");
		else
			printf("NO\n");
	}
	return 0;
}