c语言 1134

1.输入学生的成绩(0-100分)，根据分数打印等级(90分以上：A，80-89：B，70-79：C，60-69：D，60以下：E)。方法1：#include "stdio.h"main(){intgrade,x;scanf("%d",&grade);if( grade>100 || grade < 0 ){printf(":%d is error\n",grade);return ;}x=grade/10;switch (x){ case 10:case 9: printf("%d A\n",grade);break;case 8: printf("%d B\n",grade);break; case 7: printf("%d C\n",grade);break;case 6: printf("%d D\n",grade);break;default:printf("%d E\n",grade);break;}}方法2：#includevoid main(){int grade;scanf("%d",&grade);if(grade>=90&&grade<=100)printf("A");elseif(grade>=80)printf("B");elseif(grade>=70)printf("C");else if(grade>=60)printf("D");elseprintf("E");}方法3：#includevoid main(){int grade;scanf("%d",&grade);if(grade>=90&&grade<=100)printf("A");if(grade>=80&&grade<=89)printf("B");if(grade>=70&&grade<=79)printf("C");if(grade>=60&&grade<=69)printf("D");if(grade>=0&&grade<=60)printf("E");}2.在主函数中定义一个数组，输入10个数，编写一个max函数，求数组中的最大数，结果在主函数中输出。要求：数组名作实参，指针作形参.#includeint m(int *a,int n){int max=*a,i;for(i=1;ivoid main(){char str1[50]={"ab234$#DFas4"},str2[50];char *p,*q;intlen=0; p=str1; q=str2;for(;*p!='\0';p++)if(*p>='a'&&*p<='z'||*p>='A'&& *p<='Z'){ *q=*p;q++;len++;} *q=0;puts(str2);printf(“%d”,len);}4.编程打印出如下图案。 ********* ******* ***** *** *#includevoid main(){inti,j,k;for(i=4;i>=0;i--){for(j=0;j<5-i;j++)putchar(' ');for(k=0;k<2*i+1;k++)putchar('*');putchar('\n');}}5.求s=1/(1!)+1/(2!)+1/(3!)+...+1/(n!)的当n=7时的值。其中，n!是n的阶乘。结果打印出来显示9位小数。#includevoid main(){inti,n;double x=1,s=0;printf("please input n");scanf("%d",&n);for(i=1;i<=n;i++){x=x*i;s=s+1/x;}printf("s=%.9f",s);}6.编写一个求n!的函数，在main函数中由键盘输入n值，求1!+2!+3!+ …… +n!的和,并输出运算结果。方法1：#include "stdio.h