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Acwing 4301. Truncated sequence
2022-07-05 05:20:00 【hunziHang】
Given a by n A sequence of digits a1a2…an.
among , Every number is 0∼9 One of .
Please judge , Whether the sequence can be truncated from the middle into two or more non empty parts , The sum of the figures in each part is required to be equal .
for example ,350178 Can be truncated to 3 Parts of 350、17、8, And satisfy 3+5+0=1+7=8.
Input format
The first line contains an integer n.
The second line contains n A digital a1,a2,…,an, No spaces between numbers .
Output format
If you can truncate the sequence as required , The output YES
, Otherwise output NO
.
Data range
front 6 Test points meet 2≤n≤10.
All test points meet 2≤n≤100,0≤ai≤9.
Direct enumeration
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e6+100;
typedef pair<int,int> PII;
#define x first
#define y second
#define INF 0x3f3f3f3f
int dx[4]={-1,0,1,0};
int dy[4]={0,1,0,-1};
const int mod=1e9+7;
int n,m;
int a[N];
int sum;
int main()
{
string st;
cin>>n;
cin>>st;
for(int i=0;i<n;i++)
{
a[i]=st[i]-'0';
sum+=a[i];
}
for(int i=2;i<=n;i++) // Enumerate the number of segments divided into
{
if(sum%i==0) // If you can divide it, you can continue to judge
{
int f=sum/i,flag=1;
for(int j=0,s=0;j<n;j++)
{
s+=a[j];
if(s>f) // If exceeded f Then it will only get bigger and bigger , It is illegal.
{
flag=0;
break;
}
else if(s==f) // For the initial 0
s=0;
}
if(flag)
{
cout<<"YES"<<endl;
return 0;
}
}
}
cout<<"NO"<<endl;
return 0;
}
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