PAT Advanced level 1093

1093 Count PAT‘s (25)（25 分）

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.

Input Specification:

Each input file contains one test case. For each case, there is only one line giving a string of no more than 10^5^ characters containing only P, A, or T.

Output Specification:

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.

Sample Input:

APPAPT

Sample Output:

2

这题不好AC啊

/**********************
author: yomi
date: 18.6.17
ps:
不会不会 虽然乙级刷过这道题 但是现在还是不会 要多多体会这种思想
我能想到的就只有暴力循环 T^T
int len = strlen(str);
循环中一定要利用事先计算好的len
**********************/
#include <iostream>
#include <string>
#include <cstring>
#include <cstdio>
#include <stdio.h>
#include <cstdlib>
using namespace std;
const int mod = 1000000007;
int plnum[100010];
char str[100010];
int main()
{
    memset(str, 0, sizeof(str));
    memset(plnum, 0, sizeof(plnum));
    scanf("%s", str);
    int len = strlen(str);
    ///计算每一个位置的左边有多少个p，并存在数组中
    for(int i=0; i<len; i++){///如果此处不用len,而是随时计算的话，会超时
        if(i > 0){
            plnum[i] = plnum[i-1];
        }
        if(str[i] == ‘P‘){
            plnum[i]++;
        }
    }
    int ans = 0, trnum = 0;
    for(int i=len-1; i>=0; i--){
        if(str[i] == ‘T‘){
            trnum++;
        }
        else if(str[i] == ‘A‘){
            ans = (ans+plnum[i]*trnum)%mod;
        }

    }
    printf("%d\n", ans);
    return 0;
}

/***
明显下面的解法更好

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

int main()
{
    string str;
    cin >> str;
    int num_t=0, num_at=0, num_pat=0;
    ///cout << str.length() << endl;
    for(int i=str.size()-1; i>=0; i-- ){///天呐 我这就写了个size_t, 就死循环了
        if(str[i] == ‘T‘){
            num_t ++;
            ///num_t %= 1000000007;
        }
        if(str[i] == ‘A‘){
            num_at = (num_at+num_t) % 1000000007;///这里写个+=也错了 这就是优先级的事了
        }
        if(str[i] == ‘P‘){
            num_pat = (num_pat+num_at) % 1000000007;
        }
    }
    cout << num_pat;
    return 0;
}

**/

原文地址：https://www.cnblogs.com/AbsolutelyPerfect/p/9192900.html