PATB1040/A1093 有几个PAT

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.
输入格式
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.
输入格式
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.
输入样例

APPAPT

输出样例

全部AC

#include <cstdio>

#include <algorithm>

#include <cstring>

using namespace std;

const int maxn = 101000;

const int MOD = 1000000007;

int main() {

    #ifdef ONLINE_JUDGE

    #else

        freopen("1.txt", "r", stdin);

    #endif // ONLINE_JUDGE

    char str[maxn];

    scanf("%s", str);

    int len = strlen(str);

//    for(int i = 0; i < len; i++) {

//        printf("%c", str[i]);

//    }

    int times[3] = {0};

    long num = 0; //可以形成PAT的个数

    int leftNumP[maxn] = {0}, rightNumT[maxn] = {0};

    int temp = 0;

    //遍历记录每一位左边字母P的数量

    for(int i = 0; i < len; i++) {

        if(str[i] == 'P'&& temp == 0) {

            ++leftNumP[i];

            temp += 1;

        } else if(str[i] == 'P' && temp != 0) {

            leftNumP[i] = temp;

            ++leftNumP[i];

            temp += 1;

        } else leftNumP[i] = temp;

    }

    temp = 0;

    long long ans = 0;//答案

    //遍历记录每一位右边字母为T的数量

    for(int i = len; i >= 0; i--) {

        if(str[i] == 'T' && temp == 0) {

            ++rightNumT[i];

            temp += rightNumT[i];

        } else if(str[i] == 'T' && temp != 0) {

            rightNumT[i] = temp;

            ++rightNumT[i];

            temp += 1;

        } else rightNumT[i] = temp;

    }

    for(int i = 0; i < len; i++) {

        if(str[i] == 'A') {

            ans = (ans + leftNumP[i] * rightNumT[i]) % MOD;

        }

    }

//    for(int i = 0; i < len; i++) {

//        printf("leftNumP[%d]:%d   rightNumT[%d]:%d\n", i, leftNumP[i], i, rightNumT[i]);

//    }

    //直接暴力会超时

//    for(int i = 0; i < len; i++){

//        if(i != 0 && str[i] == 'A') {

//            int leftNumP = 0; //左边字母P的数量

//            int rightNumT = 0;//右边字母T的数量

//            for(int j = 0; j < i; j++) {

//                if(str[j] == 'P') leftNumP++;

//            }

//            for(int j = i; j < len; j++) {

//                if(str[j] == 'T') rightNumT++;

//            }

//            num += leftNumP * rightNumT;

//            //printf("i=%d   leftNumP:%d  rightNumT:%d num:%ld\n", i, leftNumP, rightNumT, num);

//        }

//    }

//    int ans = num % 1000000007;

    printf("%lld", ans);

    return 0;

}

巴特西

PATB1040/A1093 有几个PAT

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