C语言 45章作业

最新推荐文章于 2021-05-17 09:04:19 发布

原创最新推荐文章于 2021-05-17 09:04:19 发布 · 452 阅读

CC 4.0 BY-SA版权

本文介绍了一系列C++编程中的实用技巧与算法实现案例，包括最大值选取、平方根计算、成绩评级、收入计算、最大公约数求解等算法，并展示了字符串分析、阶乘求和、水仙花数判断等功能的实现。

#include<cstdio>
#include<cmath>
#include<cstring>

int main() {


    return 0;
}
//3

//4
int Max(int a,int b,int c) {
    return max(max(a, b), c);
}
//5
void Sqrt(int num) {
    if (num >= 0 && num <= 1000) {
        printf("%d", sqrt(num)); 
    }
    else printf("W");
}
//8
void Grade(int num) {
    if (num >= 90) {
        printf("A");
        break;
    }
    else if (num >= 80) {
        printf("B");
        break;
    }
    else if (num >= 70) {
        printf("C");
        break;
    }
    else if (num >= 60) {
        printf("D");
        break;
    }
    else printf("E");
}
//10
double income1(double x) {
    double sum = 0;
    if (x > 1000000) {
        sum = (x - 1000000)*0.01 + 100000 * 0.1 + 100000 * 0.075 + 200000 * 0.05 + 200000 * 0.03 + 400000 * 0.015;
        return sum;
    }
    if (x > 600000) {
        sum = (x - 600000)*0.015 + 100000 * 0.1 + 100000 * 0.075 + 200000 * 0.05 + 200000 * 0.03;
    }
    if (x > 400000) {
        sum = (x - 400000)*0.03 + 100000 * 0.1 + 100000 * 0.075 + 200000 * 0.05;
    }
    if (x > 200000) {
        sum = (x - 200000)*0.05 + 100000 * 0.1 + 100000 * 0.075;
        return sum;
    }
    if (x > 100000) {
        sum = (x - 100000)*0.075 + 100000 * 0.1;
        return sum;
    }
    return x * 0.1;
}
int income2(int x) {
    double sum = 0;
    int c = 0;
    if (x > 1000000)c = 1;
    else if (x > 600000)c = 2;
    else if (x > 400000)c = 3; 
    else if (x > 200000)c = 4;
    else if (x > 100000)c = 5;
    switch (c)
    {
    case 1:
        sum += (1000000 - 600000)*0.015;
    case 2:
        sum += (600000 - 400000)*0.03;
    case 3:
        sum += (400000 - 200000)*0.05;
    case 4:
        sum += (200000 - 100000)*0.075;
    case 5:
        sum += 100000*0.01;
    deflaut:
        break;
    }
    switch (c)
    {
    case 1:
        printf("%lf/n", sum + ((x-1000000)*0.01) );
        break;
    case 2:
        printf("%lf/n", sum + ((x - 600000)*0.015) );
        break;
    case 3:
        printf("%lf/n", sum + ((x - 400000)*0.03));
        break;
    case 4:
        printf("%lf/n", sum + ((x - 200000)*0.05));
        break;
    case 5:
        printf("%lf/n", sum + ((x - 100000)*0.075));
        break;
    deflaut:
        printf("%lf/n", sum + (x*0.1));
        break;
    }
}
//3
int gcd (int a, int b) {
    return (b>0) ? gcd(b, a%b) : a;
}
//4  
cahr s[105];//��ʼֵ������
void Count() {//�������� ��ָ�룿����
    scanf("%s",&s);
    int len = strlen(s);
    int eng = kon = num = oth = 0;
    for (int i = 0; i < len; i++) {
        if (s[i] == ' ') kon++;
        else if ( (s[i] <= 'z' && s[i] >= 'a') || (s[i] <= 'Z' && s[i] >= 'A') eng++;
        else if (s[i] <= '9' && s[i] >= '0')) num++;
        else oth++;
    }
    printf("%d%d%d%d%d", eng, kon, num, oth);
}
//5
void Factorial() {
    int i = 1,sum = 0��cur = 1;
    while (i <= 20) {
        cur *= i++;
        sum += cur;
    }
    printf("%d", sum);
}
//8
void Narcissistic() {
    int i, j, k, n;
    for (n = 100 ; n < 999 ; n++){
        i = n / 100;
        j = n / 10 % 10;
        k = n % 10;
        if (n == i*i*i + j*j*j + k*k*k){
            printf("%d\n", n);
        }
    }
    //printf("153��370��371��407\n");
}