题目链接codeforces
题目链接洛谷

Add One

题面翻译

称“将一个数上每一位的值$+1$”为一次操作，例如对于$93$进行一次操作后的结果为$104$。输入$n,m$，输出对$n$进行$m$次操作的结果。

题目描述

You are given an integer $ n $ . You have to apply $ m $ operations to it.

In a single operation, you must replace every digit $ d $ of the number with the decimal representation of integer $ d + 1 $ . For example, $ 1912 $ becomes $ 21023 $ after applying the operation once.

You have to find the length of $ n $ after applying $ m $ operations. Since the answer can be very large, print it modulo $ 10^9+7 $ .

输入格式

The first line contains a single integer $ t $ ( $ 1 \le t \le 2 \cdot 10^5 $ ) — the number of test cases.

The only line of each test case contains two integers $ n $ ( $ 1 \le n \le 10^9 $ ) and $ m $ ( $ 1 \le m \le 2 \cdot 10^5 $ ) — the initial number and the number of operations.

输出格式

For each test case output the length of the resulting number modulo $ 10^9+7 $ .

样例 #1

样例输入 #1

5
1912 1
5 6
999 1
88 2
12 100

样例输出 #1

5
2
6
4
2115

提示

For the first test, $ 1912 $ becomes $ 21023 $ after $ 1 $ operation which is of length $ 5 $ .

For the second test, $ 5 $ becomes $ 21 $ after $ 6 $ operations which is of length $ 2 $ .

For the third test, $ 999 $ becomes $ 101010 $ after $ 1 $ operation which is of length $ 6 $ .

For the fourth test, $ 88 $ becomes $ 1010 $ after $ 2 $ operations which is of length $ 4 $ .

我的错误代码

#include <cstdio>
#include <iostream>
#include <algorithm>

using namespace std;

const int mod = 1e9 + 7;

inline int read() {
    int s = 0, w = 1;
    char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') w = -1; c = getchar();}
    while (c >= '0' && c <= '9') s = s * 10 + c - '0', c = getchar();
    return s * w;
}

int n, m;
long long a[10][200010];
// a[初始数字][扩展次数] = 扩展后的位数

int main() {    
    for (int i = 0; i <= 9; i ++) {
        a[i][0] = 1;
    }
    
    for (int i = 0; i <= 9; i ++) { // 初始数字
        for (int j = 1; j <= 200000; j ++) { // 扩展次数
            a[i][j] = a[i][j-1];
            if (j >= 11)
                a[i][j] = (a[i][j] + a[i][j-9] - a[i][j-10] + a[i][j-10] - a[i][j-11]) % mod;
            /*
              因为数字9可以由0和1增加得到
              所以：a[i][j-9] - a[i][j-10] + a[i][j-10] - a[i][j-11]
              表示数字i扩展j次后的数字中数字9的个数。
              其中a[i][j-9] - a[i][j-10]是数字i在扩展第j-9次后
              新出现的数字1的个数；
              其中a[i][j-10] - a[i][j-11]是数字i在扩展第j-10次后
              新出现的数字0的个数；
             */
            else if (j == 10 - i || i == 9 && j == 10)
                a[i][j] = (a[i][j] + 1) % mod;
        }
    }
    
    int t = read(); 
    while (t --) {
        n = read(); m = read();
        long long sum = 0;
        while (n) {
            sum = (sum + a[n%10][m]) % mod;
            n /= 10;
        }
        printf("%lld\n", sum);
    }
    
    return 0;
}