Count PAT's (25)

最新推荐文章于 2022-10-10 11:57:33 发布

empty_coder

最新推荐文章于 2022-10-10 11:57:33 发布

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本文介绍了一种高效算法，用于计算给定字符串中特定子串‘PAT’出现的次数。通过定义p、pa、pat三个变量统计，实现了快速计算，并提供了完整的C++实现代码及思路解析。

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题目链接：https://www.nowcoder.com/pat/5/problem/4039

题目描述

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.

输入例子:

APPAPT

输出例子:

分析: 煞笔似的设计首尾两个指针来搞。。。正确姿势是，定义p、pa、pat三个统计变量搞一搞就是啦

#include <cstdio>
#include <iostream>

using namespace std;
const int N = 1e5 + 1005; 
const int MOD = 1e9 + 7;

char s[N];

void Solve() {
    //字符串最前面或最后面可能有空格
    gets(s);
    int i = -1;
    while (s[++i] == ' ');
    int p, pa, pat;
    p = pa = pat = 0;
    for (; s[i] != ' ' && s[i]; i++) {
        if (s[i] == 'P') p += 1;
        else if (s[i] == 'A') pa = (pa + p) % MOD;
        else {
            pat = (pat + pa) % MOD;
        }
    }
    printf("%d\n",pat);
}

int main() {
    Solve();
    return 0;
}