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Count PAT's (25)

[编程题]Count PAT's (25)

热度指数：2964 时间限制：C/C++ 1秒，其他语言2秒空间限制：C/C++ 32M，其他语言64M
算法知识视频讲解

The string APPAPT contains two PAT's as substrings. The first one is formed by the 2nd, the 4th, and the 6th characters,
and the second one is formed by the 3rd, the 4th, and the 6th characters.

Now given any string, you are supposed to tell the number of PAT's contained in the string.

输入描述:

Each input file contains one test case. For each case, there is only one line giving a string of no more than 10⁵
characters containing only P, A, or T.

输出描述:

For each test case, print in one line the number of PAT's contained in the string. Since the result may be a huge number, you only have to 
output the result moded by 1000000007.

示例1

输入

APPAPT

输出

算法知识视频讲解

Java

小仇Eleven

import java.util.Scanner;

public class Main {

    /**
     * @param args
     */
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String PATStr = scanner.next();
        int countP = 0;
        int result = 0;
        int countT = 0;
        for (int i = 0; i < PATStr.length(); i++) {
            if (PATStr.charAt(i) == 'T') {
                countT++;
            }
        }
        for (int i = 0; i <PATStr.length(); i++) {
            if (PATStr.charAt(i) == 'P') {
                countP++;
            }
            if (PATStr.charAt(i) == 'T') {
                countT--;
            }
            if (PATStr.charAt(i)=='A') {
                result = (result+(countP*countT)%1000000007)%1000000007;
            }
        }
        System.out.println(result);
    }

}

发表于 2018-09-05 09:48:58 回复(0)