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;
}