1.学会正确使用逻辑运算符和逻辑表达式;
2.熟练掌握if语句和switch 语句。
1.任意输入4个整数,从大到小排列输出。
2.输入一组成绩,要求输出成绩等级为A,B,C,D,E。其中90—100为A,80—89为B,70—79为C,60—69为D,60以下为E。
3.有一个函数
写一程序,输出y值。
(1)源代码
# include <stdio.h> int main(void) { int a,b,c,d,t; printf("Please enter the data(a,b,c,d):\n"); scanf("%d %d %d %d",&a,&b,&c,&d); if(a<b) { t = a; a = b; b = t; } if(a<c) { t = a; a = c; c = t; } if(a<d) { t = a; a = d; d = t; } if(b<c) { t = b; b = c; c = t; } if(b<d) { t = b; b = d; d = t; } if(c<d) { t = c; c = d; d = t; } printf("The retult is: %d %d %d %d\n",a,b,c,d); return 0; }
(2)运行结果截图
(一)源代码
# include <stdio.h> /*函数声明*/ void inputgrade(int * grade,int lg); int main(void) { int grade[10]; int lg,i; printf("How many of the data?\n"); scanf("%d",&lg); printf("Please enter the grade:\n"); inputgrade(grade,lg); for(i = 0;i<lg;++i) { if(grade[i]<=100 && grade[i]>=90) printf("The %dth degree is A!\n",i+1); else if(grade[i]<90 && grade[i]>=80) printf("The %dth degree is B!\n",i+1); else if(grade[i]<80 && grade[i]>=70) printf("The %dth degree is C!\n",i+1); else if(grade[i]<70 && grade[i]>=60) printf("The %dth degree is D!\n",i+1); else if(grade[i]<60) printf("The %dth degree is E!\n",i+1); else printf("The %dth data is wrong!"); } return 0; } /*输入数据的函数,当用户输入0时表示输入已完成。*/ void inputgrade(int * grade,int lg) { int i; for(i = 0;i<lg;++i) { printf("Please enter the %dth data:\n",i+1); scanf("%d",grade+i); } return; }
(2)运行结果截图
(1)程序框图
(2)源代码
# include <stdio.h> int main(void) { int x,y; printf("Please enter x:\n"); scanf("%d",&x); if(x<1) printf("y = %d\n",x); else if(x>=1 && x<10) printf("y = %d\n",2*x+1); else printf("y = %d\n",3*x-1); return 0; }
(3)运行结果截图
3.4 设计程序计算需要如何付钱
(1)源代码
# include <stdio.h> int main(void) { int a = 0,b = 0,c = 0,d = 0,e = 0,f = 0,g = 0,h = 0,k = 0,i,j; float mon; printf("How much the money we need?\n"); scanf("%f",&mon); /*判断需要多少张100元的钞票*/ while(mon>=100) { a++; mon = mon - 100; } /*判断需要多少张50元的钞票*/ while(mon>=50) { b++; mon = mon - 50; } /*判断需要多少张10元的钞票*/ while(mon>=10) { c++; mon = mon - 10; } /*判断需要多少张5元的钞票*/ while(mon>=5) { d++; mon = mon - 5; } /*判断需要多少张2元的钞票*/ while(mon>=2) { e++; mon = mon - 2; } /*判断需要多少张1元的钞票*/ while(mon>=1) { f++; mon = mon - 1; } /*判断需要多少张0.1元的钞票*/ while(mon>=0.1) { g++; mon = mon - 0.1; } /*判断需要多少张0.05元的钞票*/ while(mon>=0.05) { h++; mon = mon - 0.05; } /*判断需要多少张0.01元的钞票*/ while(mon>0) { k++; mon = mon - 0.01; } printf("We need:\n"); printf("100 yuan %d page(s)\n50 yuan %d page(s)\n10 yuan %d page(s)\n5 yuan %d page(s)\n2 yuan %d page(s)\n1 yuan %d page(s)\n0.1 yuan %d page(s)\n0.05 yuan %d page(s)\n0.01 yuan %d page(s)\n",a,b,c,d,e,f,g,h,k); return 0; }
(2)程序运行截图