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 yinclusively.
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
3
5
01010
6
101001
7
1101011
3
7
题意:
给定01字符串,求有多少个三元组i,j,k满足i<j<k且F(i, j)=F(j, k).其中F(x,y)是字符串下标x到y的一的个数。
下标为j的一定要是1。
思路:
先处理一下0的数量分布。比如00110就是2,0,1
假设字符串以0为开头和结尾,那么字符串便可以表示为:
a,1,b,1,c,1,d,1,e,1,f
其中字母表示0的数量。
枚举j在哪一个1上,便可以得出以下式子、
ans=a*b+ b*c+a*d+ c*d+b*e+a*f+ d*e+c*f+ e*f;
可以看出,每一个字母其实就是要与之后奇数或偶数的后缀和相乘,乘积相加就是结果。
但是这样仍然不对,因为i可k还可以指向1.
于是我们把每个字母的值加上一个1,再进行计算即可。
然而这样还是有问题。
相邻的两个字母相乘时不需要+1,于是我们再用一个for循环处理一下即可。
#include<iostream> #include<algorithm> #include<vector> #include<stack> #include<queue> #include<map> #include<set> #include<cstdio> #include<cstring> #include<cmath> #include<ctime> #define fuck(x) cout<<#x<<" = "<<x<<endl; #define ls (t<<1) #define rs ((t<<1)+1) using namespace std; typedef long long ll; typedef unsigned long long ull; const int maxn = 200086; const int inf = 2.1e9; const ll Inf = 999999999999999999; const int mod = 1000000007; const double eps = 1e-6; const double pi = acos(-1); char s[maxn]; vector<ll>v; ll sum[maxn]; int main() { int T; scanf("%d",&T); while(T--){int len; scanf("%d%s",&len,s); memset(sum,0,sizeof(sum)); v.clear(); ll ans=1; for(int i=0;i<len;i++){ if(s[i]=='0'){ ans++; } else{ v.push_back(ans); ans=1; } } v.push_back(ans); int sz=v.size(); for(int i=sz-1;i>=0;i--){ sum[i]=v[i]+sum[i+2]; } ans=0; for(int i=0;i<sz;i++){ ans+=v[i]*sum[i+1]; } for(int i=0;i<sz-1;i++){ ans-=v[i]*v[i+1]-(v[i]-1)*(v[i+1]-1); } printf("%lld ",ans); } return 0; }