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  • 复利计算3.0

    #include<stdio.h>
    #include<math.h>
    
    double danli(double i,int N,double P)
    {
        double nlx,F;
        nlx = P*i;
        F = nlx*N + P;
        return F;
    
    }
    
    double fuli(double i,int W,int N,double P)
    {
        double F;
        i=pow(1+i/W,W)-1;
        F=P*pow((i+1),N);
        return F;
    }
    
    double gusuan(double i,int W,int N,double F)
    {
        double P;
        i = pow(1+i/W,W)-1;
        P = F/pow((i+1),N);
        return P;
    }
    double shijian(double P,double i,double F)
    {
        double N,x;
        x=F/P;
        N=(log(x))/(log(i+1));
        return N;
    
    }
    double nianxian(double P,double N,double F)
    {
        double i;
         //i = W*(pow(pow(F/P,1.0/N),1.0/W)-1.0);
         i = pow(F/P,1.0/N)-1.0;
         return i;
    }
    void main()
    {
        double P,F1,F,i; //P是本金,F是收益,i年利率
        int N,W,ch2; //是期限
        while(1)
        {
        printf("		***************欢迎使用复利计算器*************
    
    ");
        printf("		|     -----1--------为复利---------------    |
    
    ");
        printf("		|     -----2--------为单利---------------    |
    
    ");
        printf("		|     -----3-------为估算----------------    |
    
    ");
        printf("		|     -----4-------为年限----------------    |
    
    ");
        printf("		|     -----5-------为年利率--------------    |
    
    ");
        printf("		|     -----0-------为退出----------------    |
    
    ");
        printf("请选择
    ");
        scanf("%d",&ch2);//控制选择
        switch(ch2){
        case 1:
            printf("请输入本金:
    ");
            scanf("%lf",&P);
            printf("请输入年利率
    ");
            scanf("%lf",&i);
            printf("请输入年限:
    ");
            scanf("%d",&N);
            printf("请输入复利次数:
    ");
            scanf("%d",&W);
            F1 = fuli(i,W,N,P); break;
        case 2:
            printf("请输入本金:
    ");
            scanf("%lf",&P);
            printf("请输入年利率
    ");
            scanf("%lf",&i);
            printf("请输入年限:
    ");
            scanf("%d",&N);
            F1 = danli(i,N,P); break;
        case 3:
            printf("请输入本利
    ");
            scanf("%lf",&F);
            printf("请输入年利率
    ");
            scanf("%lf",&i);
            printf("请输入年限:
    ");
            scanf("%d",&N);
            printf("请输入复利次数:
    ");
            scanf("%d",&W);
            F1 = gusuan(i,W,N,F);
            break;
            
        case 4:
            printf("请输入本金
    ");
            scanf("%lf",&P);
            printf("请输入年利率
    ");
            scanf("%lf",&i);
             printf("请输入本利
    ");
            scanf("%lf",&F);
            F1=shijian(P,i,F);
            break;
        case 5:
    
            printf("请输入本金:
    ");
             scanf("%lf",&P);
             printf("请输入本利:
    ");
             scanf("%lf",&F);
             printf("请输入年限
    ");
             scanf("%d",&N);
             
         
            F1=nianxian(P,N,F);
            
    //    i=pow((F/P),(1.0/N))-1;
            break;
    
        case 0:break;
        }
        if(ch2==0)
            break;
        printf("结果为:%.2lf
    ",F1);
        }
    }     
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  • 原文地址:https://www.cnblogs.com/caicaihong/p/5283044.html
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