标签:core oat 语言 break 输入 amp 没有 scanf char
#include<stdio.h> void zy1() { int a; printf("五级制成绩"); char score; scanf("%c",&score); switch (score) { case ‘A‘: printf("90~100"); break; case ‘B‘: printf("80~89"); break; case ‘C‘: printf("70~79"); break; case ‘D‘: printf("60~69"); break; case ‘E‘: printf("0~59"); break; } } void zy2() { int a; printf("请输入题目序号1-9:\n\t1.C语言基本数据类型及其占用的内存空间大小分别是:\n\t2.算数运算符有:+-*/% ++ --\n\t3.关系运算符有:> < == >= <= != <>\n\t4.逻辑运算符的优先顺序是:! && ||\n\t5.赋值运算符有:= += -= *= /= %=\n\t下面代码的输出结果是什么?请解释原因。\n\t6.\n\tchar c=132;\n\t printf(“%d”,c);\n\t7.\n\tint a=-7;\n\t printf(“%d”,a>>1);\n\t8.\n\tint a=7;\n\tint b=8;\n\tprintf(“%d”,a&b);\n\t9.\n\tint a=7;\n\tint b=8;\n\tprintf(“%d”,a^b);\n\n请选择小题"); scanf("%d",&a); switch(a) { case 1: printf("1.C语言基本数据类型及其占用的内存空间大小分别是:int:4字节,short[int]:2字节,long[int]:4字节,unsigned[int]:4字节,unsigned short[int]:2字节,unsigned long[int]:4字节,char:1字节,float:4字节,double:8字节。"); break; case 2: printf("2.算数运算符有:+,?,*,/,%,++,??。"); break; case 3: printf("3.关系运算符有:>,<,==,>=,<=,!=,<>。"); break; case 4: printf("4.逻辑运算符的有限顺序是:!,&&,||。"); break; case 5: printf("5.赋值运算符有:=,+=,-=,*=,/=,%=。"); break; case 6: printf("6.char c=132;printf(\"%d\",c);解:char的范围为-127~128,132超出范围,132变为二进制是10000100,因为在计算机中0为正1为负所以该数为负数,而负数需要取反加一,所以该数为-01111011+1=-01111100,输出结果为-124。"); break; case 7: printf("7.int a=7;printf(\"%d\",a?1);\n解:7:00000000 00000000 00000000 00000111\n-7:11111111 11111111 11111111 11111001(取反加一)\na>>1:111111111 11111111 11111111 1111100(右移一位)\n-00000000 00000000 00000000 00000100(取反进一)\n所以向右移1位为-4."); break; case 8: printf("8.int a=7;int a=8;printf(\"%d\",a&b);\n解:a=7:00000000 00000000 00000000 00000111\nb=8:00000000 00000000 00000000 00001000\n因为没有相同为1的项\n所以输出值为0."); break; case 9: printf("9.int a=7;int b=8;printf(\"%d\",a^b);\n解:a=7:00000000 00000000 00000000 00000111\nb=8:00000000 00000000 00000000 00001000\n00000000 00000000 00000000 00001111\n所以输出值为15.\n"); break; } } void zy3() { int a; printf("倒三角形"); printf("**********\n"); printf("*********\n"); printf("********\n"); printf("*******\n"); printf("******\n"); printf("*****\n"); printf("****\n"); printf("***\n"); printf("**\n"); printf("*\n"); } void zy4() { int a; printf("猜数字"); printf("请猜一个整数\n你的猜测是:"); int predefined=rand()%100; int a; scanf("%d",&a); int b; for(b=1;b<=10;b++) { if(predefined==a) { printf("恭喜你猜对啦"); } else if(predefined<a) { printf("猜大了再猜:"); scanf("%d",&a); } else { printf("猜小了再猜:"); scanf("%d",&a) ; } //scanf("%d",&a); } } void zy5() { int a; printf("最大公倍数和最小公约数"); { int a,b,i,min,max; scanf("%d%d",&a,&b); max = 1; for(i=2;i<=a;i++) { if(a%i==0 && b%i==0) max = i; } min = a*b; for(i=a*b-1;i>=a;i--) { if(i%a==0 && i%b==0) min = i; } printf("%d\n%d\n",max,min); return 0; } int choice; printf( "作业及答案查询系统\n"); printf( "1.第一次作业\n"); printf( "2.第二次作业\n"); printf( "3.第三次作业\n"); printf( "4.第四次作业\n"); printf( "5.第五次作业\n"); printf( "请选择:"); scanf("%d",&choice); switch(choice) { case 1: zy1(); break; case 2: zy2(); break; case 3: zy3(); break; case 4: zy4(); break; case 5: zy5(); break; } }
标签:core oat 语言 break 输入 amp 没有 scanf char
原文地址:http://www.cnblogs.com/zhf-9747/p/6029862.html