CodeForces 1058C C. Vasya and Golden Ticket

C. Vasya and Golden Ticket

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

Recently Vasya found a golden ticket — a sequence which consists of n digits a1a2…an. Vasya considers a ticket to be lucky if it can be divided into two or more non-intersecting segments with equal sums. For example, ticket 350178 is lucky since it can be divided into three segments 350, 17 and 8: 3+5+0=1+7=8. Note that each digit of sequence should belong to exactly one segment.

Help Vasya! Tell him if the golden ticket he found is lucky or not.

Input

The first line contains one integer n (2≤n≤100) — the number of digits in the ticket.

The second line contains n digits a1a2…an (0≤ai≤9) — the golden ticket. Digits are printed without spaces.

Output

If the golden ticket is lucky then print “YES”, otherwise print “NO” (both case insensitive).

Examples

inputCopy

5

73452

outputCopy

YES

inputCopy

4

1248

outputCopy

NO

Note

In the first example the ticket can be divided into 7, 34 and 52: 7=3+4=5+2.

In the second example it is impossible to divide ticket into segments with equal sum.

这个题求插板之后分成的区间，能否使区间和相等，我第一时间是想2分，但是后来发现实现不了。我跟队友同时开题，我想到了他们的区间和一定是区间总和的因子，然后开始做。最后卡在了55组数，队友直接全部暴力枚举过了。可以研究我的代码，要是过了给我留个言。

不AC代码

#include<cstring>
#include<iostream>
#include<cmath>
#include<set>
#include<cstdio>
#include<algorithm>
using namespace std;
void fj(int a);
    int a[120],rq[1000],num;
    bool flag;
    char c;
    int sum1=0,w;
int main()
{
    rq[0]=1;
    int n,i,sum=0;
    cin>>n;
    for(i=1;i<=n;i++)
    {
        scanf(" %c",&c);
        a[i]=c-'0';
        sum=sum+a[i];
    }
    if(sum==0) {
        cout<<"YES"<<endl;
        return 0;
    }
    fj(sum);
    sort(rq,rq+num+1);
    //for(int k=0;k<=num;k++) cout<<rq[k]<<endl;
    for(int k=0;k<=num;k++)
    {
        sum1=0;
        flag=1;
        for(i=1;i<=n;i++)
        {
           sum1+=a[i];
           if(sum1>rq[k])
           {
               flag=0;
               break;
           }
           else if(sum1==rq[k])
           {
               sum1=0;
           }
        }
        if(sum1!=0) flag=0;
        if(flag==1) break;
    }
    if(flag==1) cout<<"YES"<<endl;
    else cout<<"NO"<<endl;
return 0;
}
void fj(int a)
{
    int k=1;
    for(int i=2;i<a;i++)
    {
        if(a%i==0) rq[k++]=i;
    }
    num=k-1;
}

AC代码

#include<bits/stdc++.h>
using namespace std;
char a[105];
int n,sum=0;
bool judge(){
    bool flag=0;
    for(int i=0;i<=sum/2;i++){
        int s=0;
        if(flag) break;
        for(int j=0;j<n;j++)
            {   s+=a[j]-'0';
                if((a[j]-'0')>i||s>i) break;

                if(s==i) {
                    flag=1;
                    s=0;

                }

            }
        if(s!=0) flag=0;
    }
    return flag;
}
int main(){
    cin>>n;
    scanf("%s",a);
    for(int i=0;i<n;i++)
            sum+=(a[i]-'0');
    if(judge()) cout<<"YES";
    else cout<<"NO";

    }