数位DP

数位DP - AcWing 338 - 计数问题

注意前导0的影响

#include <bits/stdc++.h>
using namespace std;

int a, b;
int num[10];
int dp[10][10]; // 当前填位i,tar数已经出现的次数j 

int dfs(int i, int j, int flag, int first, int tar){
	if(!i) return j;
	if(!flag && !first && dp[i][j] != -1) return dp[i][j];
	int ans = 0;
	int max_num = flag ? num[i] : 9;
	for(int v = 0; v <= max_num; ++v){
		ans += dfs(i-1, j + (!(first && (!v)) && (v==tar)), flag && (v==num[i]), first && (!v), tar);
	}
	if(!flag && !first) dp[i][j] = ans;
	return ans;
}

void solve(int x, int y){
	if(x > y) swap(x, y);
	int ans[10];
	num[0] = 0;
	do{
		num[++num[0]] = y % 10;
		y /= 10;
	}while(y);
	for(int tar = 0; tar < 10; ++tar){
		memset(dp, -1, sizeof dp);
		ans[tar] = dfs(num[0], 0, 1, 1, tar) ;
	}

	--x;
	num[0] = 0;
	do{
		num[++num[0]] = x % 10;
		x /= 10;
	}while(x);
	for(int tar = 0; tar < 10; ++tar){
		memset(dp, -1, sizeof dp);
		ans[tar] -= dfs(num[0], 0, 1, 1, tar);
	}

	for(int i = 0; i < 10; ++i){
		printf("%d ", ans[i]);
	}
	putchar('
');
}

int main(){
	while(scanf("%d%d", &a, &b) && (a || b)){
		solve(a, b);
	}
	return 0;
}