Good Subarrays（思维）

Good Subarrays

题意：找给定串中满足\(\sum_{i=l}^{r}a_{i}=\left ( r-l+1\right )\)的子串的个数

题解：式子转换：

\(\sum_{i=l}^{r}a_{i}=\left ( r-l+1\right )\)连边同时减去\(\left ( r-l+1\right )\)

得：\(\sum_{i=l}^{r}\left ( a_{i}-1\right )=0\)；所以只需要找区间和为\(0\)得个数

对于前缀和：如果\(sum\left [ i\right ]==sum\left [ j\right ]\)，则\(\left ( i,j\right ]\)之间的子串之和为\(0\)

AC_Code:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
#define endl '\n'
const int inf=0x3f3f3f3f;
const int maxn=1e5+10;
const int maxm=5e5+10;
const int mod=1e9+7;

map<ll,ll>mp;
string s;
int n;
ll sum[maxn],ans;
int main()
{
    int t; cin>>t;
    while( t-- ){
        cin>>n>>s;
        sum[0]=0; ans=0; mp.clear();

        mp[0]++;//1-1=0,某个位置为1它自己就可以
        for(int i=0;i<n;i++){
            sum[i+1]=sum[i]+(s[i]-'0')-1;
            mp[sum[i+1]]++;
        }
        for(int i=0;i<=n;i++){
            if( mp[sum[i]] ) ans+=(mp[sum[i]])*(mp[sum[i]]-1)/2;
            mp[sum[i]]=0;
        }
        cout<<ans<<endl;
    }
    return 0;
}