C语言学习第四五章

谭浩强版课后习题

#include<cstdio>#include<cmath>#include<cstring>int main() {    return 0;}//3//4int Max(int a,int b,int c) {    return max(max(a, b), c);}//5void Sqrt(int num) {    if (num >= 0 && num <= 1000) {        printf("%d", sqrt(num));     }    else printf("W");}//8void 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");}//10double 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;    }}//3int gcd (int a, int b) {    return (b>0) ? gcd(b, a%b) : a;}//4  char s[105];void Count() {    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);}//5void Factorial() {    int i = 1,sum = 0，cur = 1;    while (i <= 20) {        cur *= i++;        sum += cur;    }    printf("%d", sum);}//8void 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");}