紫书——Digit Generator UVA - 1583

题解:

改题目要求求出最小生成元,那么我们由1到100000打表出他们的最小生成元,在printf就ok了

#include <cstdio>
#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
#include <map>

using namespace std;

int biao[100010];

int main() {
	//freopen("in.txt","r",stdin);
	//freopen("output.txt","w",stdout);
	
	for(int i = 1; i < 100000; i++){
		int tmp = i; int sum = tmp;
		while(tmp > 0){
			sum += tmp%10;
			tmp /= 10;
		}
		if(sum <= 100000 && !biao[sum])	biao[sum] = i;	
	}
	
	int n;
	scanf("%d",&n);
	for(int i = 0; i < n; i++){
		int tmp;
		scanf("%d",&tmp);
		if(biao[tmp]) printf("%d\n",biao[tmp]);
		else printf("0\n");
	}

	return 0;
}

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转载自blog.csdn.net/a673953508/article/details/80105246