2020计算机C语言

文件后缀“.cpp” 提交选择“G++”编译器

8

 1 #include<stdio.h>
 2 int num[100];
 3 int main()
 4 {
 5     int x;
 6     int cnt = 0; 
 7     scanf("%d", &x);
 8     while(x) {
 9         num[++cnt] = x % 10;
10         x = x / 10;
11     }
12     if(!cnt) printf("0
");
13     else {
14         for(int i = cnt; i>=1; i--)
15             printf("%d
", num[i]);
16     }
17     return 0; 
18 }
View Code

9

1 #include<stdio.h>
2 int main()
3 {
4     int x;
5     scanf("%d", &x);
6     printf("%c
", x);
7     return 0;
8 }
View Code

10

#include<stdio.h>
int main()
{
    int x; 
    scanf("%d",&x);
    printf("x=%d,x=%o,x=%x
", x, x, x);
    return 0;
}
// -1
// x=-1,x=37777777777, x=ffffffff
View Code

11

#include<stdio.h>
int main()
{
    int x, y, ans;
    char op;
    scanf("%d %c %d", &x, &op, &y);
    if(op=='+')
        ans = x + y;
    else if(op=='-')
        ans = x - y;
    else if(op=='/')
        ans = x / y;
    else
        ans = x % y;
    printf("%d
", ans);
    return 0;
}
View Code

12

#include<stdio.h>
int fun(int x, char op, int y) {
    int ans;
    if(op=='+')
        ans = x + y;
    else if(op=='-')
        ans = x - y;
    else if(op=='/')
        ans = x / y;
    else
        ans = x % y;
    return ans;
}
int main()
{
    int x, y;
    char op;
    scanf("%d %d %c",&x, &y, &op);
    if(op=='/' && y==0)
        printf("Go to hell!
");
    printf("%d
", fun(x, op, y));
    return 0;
}
View Code

13

#include<stdio.h>
int main()
{
    int op; 
    double d, ans;
    scanf("%d %lf", &op, &d);
    if(op==1) {
        ans = (d - 32) * 5 / 9;
        printf("The Centigrade is %.2lf
", ans);
    }
    else {
        ans = (d * 9 / 5) + 32;
        printf("The Fahrenheit is %.2lf
", ans);
    }
    return 0;
}
View Code

16

#include<stdio.h>
int main()
{
    char op;
    op = getchar();
    if(op>='a' && op<='z')
        op = op - 'a' + 'A';
    else if(op>='A' && op<='Z')
        op = op - 'A' + 'a';
    printf("%c
", op);
    return 0;
}
View Code

19

#include<stdio.h>
#include<math.h>
int main()
{
    int a, b, c;
    double x1, x2;
    scanf("%d %d %d", &a, &b, &c);
    if(a==0&&b==0)
        printf("Input error!
");
    else if(a==0){
        x1 = - double(b) / c;
        printf("x=%.6f
", x1);
    } 
    else {
        double t = b * b - 4 * a * c;
        if(t==0)
            printf("x1=x2=%.6f
",-double(b)/(2*a));
        else if(t>0){
            x1 = (-b + sqrt(t)) / (2 * a);
            x2 = (-b - sqrt(t)) / (2 * a);
            printf("x1=%.6f
x2=%.6f
",x1, x2);
        }  
        else {
            x1 = -double(b) / (2 * a);
            x2 = sqrt(-t) / (2 * a);
            printf("x1=%.6f%+.6fi
", x1, x2);
            printf("x2=%.6f%+.6fi
", x1, -x2);
        }
    }
    return 0;
}
View Code

20

#include<stdio.h>
#include<math.h>
int main()
{
    int x, y;
    while(scanf("%d %d", &x, &y)!=EOF) {
        // 分针走的时候 时针也走
        double x_pos = (x%12) * 5  + double(y) / 60.0 * 5;
        // 把整个表分成60份:  (x%12) * 5 -- 时针本来的位置; double(y) / 60.0 * 5 -- 因为分针,时针走的位置 
        double ans = fabs(x_pos - y) < 30.0 ? fabs(x_pos - y) : 60.0 - fabs(x_pos - y);
        // 求最小夹角
        printf("At %02d:%02d the angle is %.1lf degrees.
", x, y, ans * 6.0);
        // ans * 6.0 一份是6度
    }
    return 0;
}
View Code

 38

#include <stdio.h>
#include <math.h>
#include <string.h>
char s[100];
int n;
int get_point(char* s) {
    int len = strlen(s);
    for(int i=0; i < len; i++)
        if(s[i] == '.')
            return i;
    // 整数末尾添加.  
    s[len] = '.';
    return len;
}
main() 
{
    char op;
    scanf("%s %c %d", s, &op, &n);
    int point = get_point(s);
    int num = fabs(n);
    int i;
    for(i = 0; i<num; i++) {
        if(n>0) {
            s[point] = s[point+1];
            if(s[point] == '')
                s[point] = '0';
            s[point+1] = '.';
            point += 1;
        }
        else{
            s[point] = s[point-1];
            s[point-1] = '.';
            if(point-1==0)  {
                char s0[100] = "0";
                strcat(s0, s);
                strcpy(s, s0);
            }
            else      point -= 1;
        }
    }
    for(int i=point+1; i<=point+8; i++) 
        if(s[i]=='')
            s[i] = '0';
    s[point+9] = '';
    printf("%s
", s);
    return 0;
}
View Code

40

#include<bits/stdc++.h>  
using namespace std;  
int a[100], b[100];  
char c1[100], c2[100];  
int fun(char* c, int* a, int n) {  
    int num = 2;  
    for(int i=0; i<n; i++) {  
        if(c[i]>='0'&&c[i]<='9')  
            a[i] = c[i] - '0';  
        else  
            a[i] = c[i] - 'A' + 10;  
        num = max(num, a[i]+1);  
    }  
    return num;  
}  
int get(int t, int* a, int n) {  
    int ans = 0;  
    for(int i=0; i<n; i++)  
        ans = ans * t + a[i];  
    return ans;  
}  
int main()  
{  
    scanf("%s %s", c1, c2);  
    int n1 = strlen(c1);  
    int n2 = strlen(c2);  
    int x1 = fun(c1, a, n1);  
    int x2 = fun(c2, b, n2);  
    bool isok=false;  
    for(int i=x1; i<=36&&!isok; i++) {  
        int t1 = get(i, a, n1);  
        for(int j=x2; j<=36&&!isok; j++)  {  
            int t2 = get(j, b, n2);  
            if(t1 == t2) {  
                isok = true;  
                x1 = i;  
                x2 = j;  
            }  
        }  
    }  
    if(!isok)   printf("%s is not equal to %s in any base 2..36
", c1, c2);  
    else        printf("%s (base %d) = %s (base %d)
", c1, x1, c2, x2);  
    return 0;  
}  
View Code
原文地址:https://www.cnblogs.com/xidian-mao/p/14008263.html