N - Modulo Sum（01背包）

桃扇骨 2022-02-23 04:28 177阅读 0赞

### 题目描述： ###

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.  
Examples  
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.

题意：  
从给出的序列中，能否选出几个数整除m。

分析：  
如果ｎ〉ｍ则一定可以。　  
ｎ＜＝ｍ的话，就需要通过０１背包操作了。  
基本思路还是用前ｉ个数能组成哪些数，　并把那些数统统标记为１.  
因为这个题的数据较大， 所以我们可以通过mod m 来使数据变小。  
这样的话， 我们只需要最后判断一下，dp\[i\]\[0\]是否被标记过。 如果被标记过， 则表明可以整除。

#include"stdio.h"
    #include"string.h"
    #include"functional"
    #include"iostream"
    #include"algorithm"
    using namespace std;
    typedef long long ll;
    #define scanll(a,b) scanf("%I64d%I64d",&a,&b);
    #define scanl(a) scanf("%I64d",&a);
    #define scanff(a,b) scanf("%lf%lf",&a,&b);
    #define scan1f(a) scanf("%lf",&a);
    #define prinll(a,b) printf("%I64d %I64d",a,b);
    #define prinl(a) printf("%I64d",a);
    #define printff(a,b) printf("%lf %lf",a,b);
    #define printlf(a) printf("%lf",a);
    #define OK printf("\n");
    #define MAXSIXE 1001000
    #define MAX 1000000010
    #define INF 11001010
    ll a[MAXSIXE];
    ll dp[1010][1010];
    ll cnt[1010];
    int main()
    {
        ll n,m;
        while(~scanf("%I64d%I64d",&n,&m))
        {
            ll sum=0;
            for(ll i=1; i<=n; i++)
            {
                scanl(a[i]);
                a[i]=a[i]%m;
            }
            int MARK=0;
            if(n>m)
            {
                printf("YES\n");
                return 0;
            }
            memset(dp,0,sizeof(dp));
            for(ll i=1; i<=n; i++)
            {
                dp[i][a[i]]=1;//这里 
                for(ll j=0; j<m; j++)
                {
                    if(dp[i-1][j])
                    {
                        dp[i][j]=1;
                        dp[i][(j+a[i])%m]=1;
                    }
                }
            }
            for(ll i=1;i<=n;i++)
                if(dp[i][0])  MARK=1;
            if(MARK == 1)
                printf("YES\n");
            else
                printf("NO\n");
        }
    }

参考博客：[https://blog.csdn.net/loveyou11111111/article/details/48595091][https_blog.csdn.net_loveyou11111111_article_details_48595091]  
[https://blog.csdn.net/h1021456873/article/details/48608699][https_blog.csdn.net_h1021456873_article_details_48608699]

[https_blog.csdn.net_loveyou11111111_article_details_48595091]: https://blog.csdn.net/loveyou11111111/article/details/48595091
[https_blog.csdn.net_h1021456873_article_details_48608699]: https://blog.csdn.net/h1021456873/article/details/48608699