5.12打卡

 二、思路设计

 三、程序流程图

 四、代码实现

#include<bits/stdc++.h>
using namespace std;
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;
                    
                    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);
                    
                    if(result > max){
                        max = result;
                        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息合计：%0.2f\n", result); 
    
}