#include <stdio.h> int main() { double a[6] = { 0,100000,200000,400000,600000,1000000 }, t[6] = { 0.1,0.075,0.05,0.03,0.015,0.01 },o=0; int n,p; scanf("%d", &p); for (n = 0; p > a[n+1]; n++) { o = o + (a[n + 1] - a[n]) * t[n]; if (n == 4) { n++; break; } } if (p < a[1])printf("%d", (int)(p * t[0])); else printf("%d", (int)(o + (p - a[n]) * t[n])); return 0; } 用循环搞了一下
廖智贤 2020-10-30 12:16:52 |
哈哈,循环是真的不错,简化了代码行数
廖智贤 2020-10-30 12:17:43 |
实现了我当初想要用循环的想法,哈哈,谢谢啦