Find Q
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/131072 K (Java/Others)Total Submission(s): 742 Accepted Submission(s): 364
Problem Description
Byteasar is addicted to the English letter 'q'. Now he comes across a string
S
consisting of lowercase English letters.
He wants to find all the continous substrings of S, which only contain the letter 'q'. But this string is really really long, so could you please write a program to help him?
He wants to find all the continous substrings of S, which only contain the letter 'q'. But this string is really really long, so could you please write a program to help him?
Input
The first line of the input contains an integer
T(1≤T≤10),
denoting the number of test cases.
In each test case, there is a string S, it is guaranteed that S only contains lowercase letters and the length of S is no more than 100000.
In each test case, there is a string S, it is guaranteed that S only contains lowercase letters and the length of S is no more than 100000.
Output
For each test case, print a line with an integer, denoting the number of continous substrings of
S,
which only contain the letter 'q'.
Sample Input
2 qoder quailtyqqq
Sample Output
1 7
很简单的题,有一个坑要用存放答案的值,要用long long才能过
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
char str[100004];
int main()
{
int N,i;
scanf("%d",&N);
while(N--)
{
scanf("%s",str);
int len = strlen(str);
long long ans = 0,num = 0;
for(i = 0; i < len; i++)
{
if(str[i]=='q')
{
num++;
}
else
{
ans = ans + (num*(num+1))/2;
num = 0;
}
}
ans = ans + (num*(num+1))/2;
printf("%lld\n",ans);
}
return 0;
}
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
char str[100004];
int main()
{
int N,i;
scanf("%d",&N);
while(N--)
{
scanf("%s",str);
int len = strlen(str);
long long ans = 0,num = 0;
for(i = 0; i < len; i++)
{
if(str[i]=='q')
{
num++;
}
else
{
ans = ans + (num*(num+1))/2;
num = 0;
}
}
ans = ans + (num*(num+1))/2;
printf("%lld\n",ans);
}
return 0;
}