AtCoder Beginner Contest 135 D - Digits Parade

链接地址：https://atcoder.jp/contests/abc135/tasks/abc135_d

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 400 points

Problem Statement

Given is a string S. Each character in S is either a digit (0, ..., 9) or ?.

Among the integers obtained by replacing each occurrence of ? with a digit, how many have a remainder of 55 when divided by 13? An integer may begin with 0.

Since the answer can be enormous, print the count modulo 109+7109+7.

Constraints

• SS is a string consisting of digits (0, ..., 9) and ?.
• 1≤|S|≤10^5

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Input

Input is given from Standard Input in the following format:

S

Output

Print the number of integers satisfying the condition, modulo 109+7.

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Sample Input 1 Copy

??2??5

Sample Output 1 Copy

768

For example, 482305,002865,482305,002865, and 972665972665 satisfy the condition.

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Sample Input 2 Copy

?44

Sample Output 2 Copy

1

Only 044044 satisfies the condition.

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Sample Input 3 Copy

7?4

Sample Output 3 Copy

0

We may not be able to produce an integer satisfying the condition.

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Sample Input 4 Copy

?6?42???8??2??06243????9??3???7258??5??7???????774????4?1??17???9?5?70???76???

Sample Output 4 Copy

153716888

题意：

给你一个由数字（0,9）和？组成的字符串，问你有几种情况该字符串模13余5，可以有前缀零，如果答案过大，则对1e9+7取模。

思路：dp

dp[100005][13],dp[i][j]表示到第i个字符为止，除以13余j(j表示的是到i-1为止模13余j)有多少中情况

若当前字符为数字，则递推公式dp[i][(j*10+s[i])%13] = (dp[i][(j*10+s[i])%13] + dp[i-1][j])%mod;

若当前字符为？，则递推公式dp[i][(j*10+k)%13] = (dp[i][(j*10+k)%13] + dp[i-1][j])%mod;  (k从0到9，假设当前字符为k)

该过程就是模拟某个数模13余几的过程，比如：567%13 -> 5%13=5 , (5*10+6)%13=4 , (4*10+7)%13=8

以下代码为atcoder某位大佬（刷新下发现忘了是哪位大佬了）的代码

#include 
using namespace std;
const int mod = 1e9+7;
char s[100005];
int dp[100005][13];
int main(){
    scanf("%s", s+1);
    int n = strlen(s+1);
    dp[0][0] = 1;
    for(int i=1;i<=n;i++){
        if(s[i] != '?'){
            s[i]-='0';
            for(int j=0;j<13;j++){
                dp[i][(j*10+s[i])%13] = (dp[i][(j*10+s[i])%13] + dp[i-1][j])%mod;
            }
        }
        else{
            for(int k=0;k<10;k++){
                for(int j=0;j<13;j++){
                    dp[i][(j*10+k)%13] = (dp[i][(j*10+k)%13] + dp[i-1][j])%mod;
                }   
            }
        }
    }
    printf("%d\n", dp[n][5]);
}