###__DP大法__
```c++
评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 39652 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 39648 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 39648 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 39648 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 39648 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 39648 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 39648 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 39648 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 39648 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 39648 KiB, score = 10
Accepted, time = 0 ms, mem = 39652 KiB, score = 100
代码
#include <cstdio>
int dp[10000001],n,x,sum = 0;
int main() {
scanf("%d%d",&n,&x);
dp[x] = 1;
for (int i = 1;i <= n;i++) {
dp[i] = dp[i%10];
if (i/10) dp[i] += dp[i/10];
sum += dp[i];
}
printf("%d",sum);
return 0;
}
```