题解 | #零钱兑换#
零钱兑换
https://ac.nowcoder.com/acm/problem/22197
#include <stdio.h>
using namespace std;
int main()
{
int n;
scanf("%d", &n);
int result = n/2+1;
for(int i = 1; i <= n/5;i++)/*n/5代表最多含有多少个5,以进行循环*/
{
int num_5 = n - i*5;/*num_5代表减去多少个数字5后剩下的数字*/
result += num_5/2+1;
}
printf("%d",result);
return 0;
}
