Codeforces gym 101350A dp

Sherlock Bones

time limit per test
1.5 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

The great dog detective Sherlock Bones is on the verge of a new discovery. But for this problem, he needs the help of his most trusted advisor -you- to help him fetch the answer to this case.

He is given a string of zeros and ones and length N.

Let F(x, y) equal to the number of ones in the string between indices x and y inclusively.

Your task is to help Sherlock Bones find the number of ways to choose indices (i, j, k) such that i < j < k, sj is
equal to 1, and F(i, j) is
equal to F(j, k).

Input

The first line of input is T – the number of test cases.

The first line of each test case is an integer N (3 ≤ N ≤ 2 × 105).

The second line is a string of zeros and ones of length N.

Output

For each test case, output a line containing a single integer- the number of ways to choose indices (i, j, k).

Example

input

3
5
01010
6
101001
7
1101011

output

2
3
7

题意：定义f(i,j)为[i,j]中1的个数  找出有多少三元组  f(i,j)=f(j,k)且i<j<k且a[j]=1

题解：

定义dp[0][0]为不取前一位的和 mod 2=0的情况  dp[1][0]为取前一位的和  mod 2=0的情况

dp[1][0]为不取前一位的和 mod 2=1的情况  dp[1][1]为取前一位的和  mod 2=1的情况

dp搞一搞  然后去掉  100000000    000000001  1  这些情况

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
ll dp[2][2];
char s[200005];
int main(){
int t;
scanf("%d",&t);
while(t--){
int n,i,j;
scanf("%d",&n);
memset(dp,0,sizeof(dp));
scanf("%s",s+1);
for(i=1;i<=n;i++){
if(s[i]=='0'){
dp[0][0]+=dp[0][1];
dp[1][0]+=dp[1][1];
dp[0][1]++;
}
else{
dp[0][0]+=dp[0][1];
dp[1][0]+=dp[1][1];
swap(dp[1][1],dp[0][1]);
dp[1][1]++;
}
}
ll ans=dp[1][0]+dp[1][1];
for(i=1;i<=n;i++){
if(s[i]=='1'){
ans--;
int now=i-1;
while(s[now]=='0'&&now>=1){
ans--;
now--;
}
now=i+1;
while(s[now]=='0'&&now<=n){
ans--;
now++;
}
}
}
printf("%lld\n",ans);
}
return 0;
}