标签：sicily

Constraints

Time Limit: 1 secs, Memory Limit: 256 MB

Description

Given two words A and B, a word W is said to be good if it satisfies the following conditions simultaneously.

1) All letters in W are also in A.

2) No letter in W is in B.

3) W contains N letters.

Given N, A and B, your task is to find out how many different good words exist.

Input

The input begins with a line containing an integer T (T<=50), which indicates the number of test cases. The following T lines each contain an integer N (1<=N<=10000), and two words A and B. A and B only contain lowercase English letters. The length of each word will not exceed 50.

Output

For each case, output the number of different good words in a line. The answer may be very large, so just output the remainder of the answer after divided by 1007.

Sample Input

3
3 lby myf
1 ddfg ffgd
5 lby ygl

Sample Output

8
0
1

Hint

For the first test case, you can generate 8 good words: lll, llb, lbl, lbb, bll, blb, bbl, bbb.

Problem Source

SYSUCPC 2014 Preliminary (Online) Round

#include <iostream>
#include <string>
using namespace std;

int main() {

	std::ios::sync_with_stdio(false);

	int caseNum;
	cin >> caseNum;
	
	while (caseNum--) {
		int n, m = 0, ans = 1;
		string a, b;
		cin >> n >> a >> b;
		bool isOK['z' + 1];
		for (int i = 'a'; i <= 'z'; i++) isOK[i] = false;
		for (int i = a.size() - 1; i >= 0; i--) isOK[a[i]] = true;
		for (int i = b.size() - 1; i >= 0; i--) isOK[b[i]] = false;
		for (int i = 'a'; i <= 'z'; i++) if (isOK[i]) m++;
		for (int i = 0; i < n; i++, ans %= 1007) ans *= m;
		cout << ans << endl;
	}

	return 0;
}

Sicily 14261. Generating Words

标签：sicily

原文地址：http://blog.csdn.net/u012925008/article/details/44929431