编程介的小学生 2019-08-14 21:56 采纳率: 0.4%
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字符串的记数问题,怎么用C语言来写的

Problem Description
It is well known that AekdyCoin is good at string problems as well as number theory problems. When given a string s, we can write down all the non-empty prefixes of this string. For example:
s: "abab"
The prefixes are: "a", "ab", "aba", "abab"
For each prefix, we can count the times it matches in s. So we can see that prefix "a" matches twice, "ab" matches twice too, "aba" matches once, and "abab" matches once. Now you are asked to calculate the sum of the match times for all the prefixes. For "abab", it is 2 + 2 + 1 + 1 = 6.
The answer may be very large, so output the answer mod 10007.

Input
The first line is a single integer T, indicating the number of test cases.
For each case, the first line is an integer n (1 <= n <= 200000), which is the length of string s. A line follows giving the string s. The characters in the strings are all lower-case letters.

Output
For each case, output only one number: the sum of the match times for all the prefixes of s mod 10007.

Sample Input
1
4
abab

Sample Output
6

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1条回答 默认 最新

  • 你知我知皆知 2024-08-11 06:12
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    以下回答参考 皆我百晓生券券喵儿 等免费微信小程序相关内容作答,并由本人整理回复。

    #include <stdio.h>
    int main(){
    	int t,n;
    	scanf("%d",&t);
    	while(t--){
    		scanf("%d",&n);
    		char str[200005];
    		scanf("%s",str);
    		long long ans=0;
    		for(int i=0;i<n;++i){
    			if(str[i]=='a'){
    				ans+=1LL*(i+1)*i/2%10007;
    			}
    			else{
    				ans+=1LL*(n-i-1)*(n-i)/2%10007;
    			}
    		}
    		printf("%lld\n",ans%10007);
    	}
    	return 0;
    }
    
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