AtCoder Beginner Contest 104 D题题解

D - We Love ABC

题目

Problem Statement

The ABC number of a string T is the number of triples of integers (i,j,k) that satisfy all of the following conditions:

• 1≤i<j<k≤|T| (|T| is the length of T.)
• T**i= A (T**i is the i-th character of T from the beginning.)
• T**j= B
• T**k= C

For example, when T= ABCBC, there are three triples of integers (i,j,k) that satisfy the conditions: (1,2,3),(1,2,5),(1,4,5). Thus, the ABC number of T is 3.

You are given a string S. Each character of S is A, B, C or ?.

Let Q be the number of occurrences of ? in S. We can make 3Q strings by replacing each occurrence of ? in S with A, B or C. Find the sum of the ABC numbers of all these strings.

This sum can be extremely large, so print the sum modulo 109+7.

Constraints

• 3≤|S|≤105
• Each character of S is A, B, C or ?.

Input

Input is given from Standard Input in the following format:

S

Output

Print the sum of the ABC numbers of all the 3Q strings, modulo 109+7.

Sample Input 1

A??C

Sample Output 1

8

In this case, Q=2, and we can make 3Q=9 strings by by replacing each occurrence of ? with A, B or C. The ABC number of each of these strings is as follows:

• AAAC: 0
• AABC: 2
• AACC: 0
• ABAC: 1
• ABBC: 2
• ABCC: 2
• ACAC: 0
• ACBC: 1
• ACCC: 0

The sum of these is 0+2+0+1+2+2+0+1+0=8, so we print 8 modulo 109+7, that is, 8.

题意

给出一个序列，由'A','B','C'或 '?'组成。

思路

dp的思路，从长度末尾开始，如果j=3时，当前位为？表示可能有三种选择。否则只有一种选择。在j<3时，如果当前位置为ABC的正确位置或者为'?'时，ij应当加上i+1 j+1的情况。

代码

#pragma comment(linker, ¡°/STACK:1024000000,1024000000¡±)
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 0x3f3f3f3f;

#define REP(i,n) for(int i=0;i<(n);i++)

int main(){
    string S;
    cin >> S;
    int N = S.size(), MOD = 1000000007;

    long long dp[100010][4] = {};
    for(int i = N; i >= 0; --i){
        for(int j = 3; j >= 0; --j){
            if(i == N){
                dp[i][j] = (j == 3 ? 1 : 0);
            }else{
                dp[i][j] = dp[i+1][j] * (S[i] == '?' ? 3LL : 1LL);
                if(j < 3 && (S[i] == '?' || S[i] == "ABC"[j])){
                    dp[i][j] += dp[i+1][j+1];
                }
                dp[i][j] %= MOD;
            }
        }
    }
    cout << dp[0][0] << endl;
}