【codechef】 Count Substrings (大数技巧题)（二分+计数，易超

You are given a string?S?of length?N?consisting only of?0s and?1s. You are also given an integer?K.

You have to answer?Q?queries. In the?i^th?query,two integers?L_i?and?R_i?are given. Then you should print the number of substrings of?S[L,R]?which contain at most?K?0s and at most?K?1s where?S[L,R]?denotes the substring from?L^th?to?R^th?characters of the string?S.?
In other words,you have to count number of pairs?(i,j)?of integers such that?L ≤ i ≤ j ≤ R?such that no character in substring?S[i,j]?occurs more than?K?times.

Input

The first line of input contains an integer?T,denoting the number of test cases. Then?T?test cases follow.

The first line of each test case contains three space-separated integers?N,?K?and?Q?as described in the problem. The second line contains a string?S?of length?N. Then the next?Q?lines describe the query,where the?i^th?line of them contains two space-separated integers?L_i?and?R_i.

Output

For each query,print the required answer in a single line.

Constraints and Subtasks

• 1 ≤ T ≤ 10⁵
• 1 ≤ K ≤ N ≤ 10⁵
• 1 ≤ Q ≤ 10⁵
• 1 ≤ L_i?≤ R_i?≤ N

Input:
1
8 2 3
01110000
1 4
2 4
5 8

Output:
8
5
7

http://www.codechef.com/problems/STRSUB/

这题非常非常莫名其妙地WA了20次！！！

下面是这道题目的两种解法：

#include<iostream>  
#include<algorithm>  
#include<string>  
#include<map>  
#include<set>  
#include<cmath>  
#include<string.h>  
#include<stdlib.h>  
#include<cstdio>  
#define ll long long  
using namespace std; 
int main(){  
    int t;  
    cin>>t;  
    while(t--){  
        ll n,m,k,a,b;  
        string x;  
        cin>>n>>m>>k;  
        cin>>x;  
        ll y[n+5]={0};  
        ll sum[n+5]={0};
        ll c=0,d=0,r=0;  
        for(int i=0;i<n;++i){  
            while(r<n){   //这个r很好，大大缩短了时间（以后的i的r只要在前面基础上累加就行了）  
                if(x[r]=='1'&&c<m){  
                    c++;  
                    r++;  
                }  
                else if(x[r]=='0'&&d<m){  
                    d++;  
                    r++;  
                }  
                else{  
                    break;  
                }  
            }  
            y[i+1]=r;   //以每个位置作为开头时的结尾位置  
            sum[i+1]+=sum[i]+r-i; //这个位置之前(包括这个位置),所有以这些位置为起点的符合条件的子集长度和，即有多少个子集  
            if(x[i]=='1')  
                c--;  
            else  
                d--;  
        }
        while(k--){
            cin>>a>>b;  
            int left=upper_bound(y+a,y+b,b)-y; //取等于时的最后一个下标，如果没有等于的，取大于的第一个下标 
            if(y[left]>b)  //当这个下标取的是大于的第一个下标 
            	left--;
            cout<<sum[left]-sum[a-1]+(b-left)*(b-left+1)/2<<endl;//sum的差表示left前（包括）子集总数，后面可以看成C(b-left,2)+b-left(单个点也是子集)  
        }
    }
    return 0;     
}
或

int main(){
	int t;
	cin>>t;
	while(t--){
		ll n,b;
		string x;
		cin>>n>>m>>k;
		cin>>x;
		ll y[n+5]={0};    //不知道为什么写y[100005]就是超时？
		ll sum[n+5]={0};  //不知道为什么写y[100005]就是超时？
		ll c=0,r=0;
		for(int i=0;i<n;++i){
			while(r<n){   //这个r很好，大大缩短了时间（以后的i的r只要在前面基础上累加就行了）
				if(x[r]=='1'&&c<m){
					c++;
					r++;
				}
				else if(x[r]=='0'&&d<m){
					d++;
					r++;
				}
				else{
					break;
				}
			}
			y[i+1]=r;   //以每个位置作为开头时的结尾位置
			sum[i+1]+=sum[i]+r-i; //这个位置之前(包括这个位置),所有以这些位置为起点的符合条件的子集长度和，即有多少个子集
			if(x[i]=='1')
				c--;
			else
				d--;
		}
		while(k--){
			cin>>a>>b;
			ll left=a,right=b;
			ll mid;
			while(left<right){  //注意这里没有等于
				mid=(left+right)/2;
				if(y[mid]>b)  //这个地方我b写成r，WA了无数次（教训啊）
					right=mid-1;
				else
					left=mid+1;
			}
			if(y[left]>b)   //这一步很重要（为什么取left而不是mid？）
				left--;
		cout<<sum[left]-sum[a-1]+(b-left)*(b-left+1)/2<<endl;//sum的差表示left前（包括）子集总数，后面可以看成C(b-left,2)+b-left(单个点也是子集)
		}
	}
	return 0;	
}