#include <stdio.h>
#include<math.h>
int main()
{
int x1, x2, x3, x5, x8, y1, y2, y3, y5, y8;
double max = 0.0, result;
for (x8 = 0; x8 <= 2; x8++)
for (x5 = 0; x5 <= (20 - 8 * x8) / 5; x5++)
for (x3 = 0; x3 <= (20 - 8 * x8 - 5 * x5) / 3; x3++)
for (x2 = 0; x2 <= (20 - 8 * x8 - 5 * x5 - 3 * x3) / 2; x2++)
{
x1 = 20 - 8 * x8 - 5 * x5 - 3 * x3 - 2 * x2; //存款期限限定条件
//判断条件
result = 2000.0 * pow((1 + 0.0063 * 12), x1)
* pow((1 + 2 * 0.0066 * 12), x2)
* pow((1 + 3 * 0.0069 * 12), x3)
* pow((1 + 5 * 0.0075 * 12), x5)
* pow((1 + 8 * 0.0084 * 12), x8);
//y1,y2,y3,y5,y8用于记录获利最多的存放方式
if (result > max)
{
max = result; //max变量存放当前的最大值
y1 = x1;
y2 = x2;
y3 = x3;
y5 = x5;
y8 = x8;
}
}
printf("获得利息最多的存款方式为:\n");
printf("8年期限的存了%d次\n", y8);
printf("5年期限的存了%d次\n", y5);
printf("3年期限的存了%d次\n", y3);
printf("2年期限的存了%d次\n", y2);
printf("1年期限的存了%d次\n", y1);
printf("存款人最终的获得的本利合计:%0.2f\n", result);
return 0;
}
标签:max,问题,y8,x8,result,printf,y1,存钱
From: https://www.cnblogs.com/258-333/p/17352248.html