第十二届蓝桥杯省赛C&C++ 研究生组-卡片

solution1

直接依次计算，思维上最直接

#include<stdio.h>
int main(){
	int a[10], n = 1, t1, t2, flag = 1;
	for(int i = 0; i < 10; i++)
		a[i] = 2021;
	while(flag){
		t1 = n;
		while(t1){
			t2 = t1 % 10;
			a[t2]--;
			if(a[t2] < 0){
				flag = 0;
				break;
			} 
			t1 /= 10;
		}
		if(!t1 && flag) n++;
	}
	printf("%d", n - 1);
	return 0;
}

solution2

观察可知，点数最小的卡片1最先消耗完

==》问题转化为1什么时候用光

#include<stdio.h>
int main(){
	int n = 1, t1, t2, flag = 1, one = 0;
	while(flag){
		t1 = n;
		while(t1){
			t2 = t1 % 10;
			if(t2 == 1) one++;
			if(one == 2021){
				flag = 0;
				break;
			}
			t1 /= 10;
		}
		if(!t1 && flag) n++;
	}
	printf("%d", n);
	return 0;
}