写一个c 程序计算1到100的和,并改进可以计算任意2个整数之间的和.

intmain（void）｛inti,ans;For(i=1;i再答：�������ټ�һ��#include再问：лл

////////////////////下面是whileintsum=0;int_i=0;while(_i

最简单的求法就是使用等差数列的求和公式,参考程序如下：#includeintmain(void){intnumber;number=(0+100)*101/2;printf("%d\n",number

#includeintmain(){inti;doublesum=0.0;doublejiecheng(doublenumber);for(i=1.0;i

#includevoidmain(){inta,b;printf("a=");scanf("%d",&a);printf("b=");scanf("%d",&b);printf("\na+b=%d\n

intarea(intx,inty){returnx*y;}intgri(intx,inty){return2*(x+y);}

#includevoidmain(){intsum=0;for(inti=1;i

PublicSubqqqqq()DimiAsIntegerDimjAsIntegerDimsAsIntegerDimCAsBooleans=0Fori=2To100C=FalseForj=2Toi-1

dimiaslongdimsumaslongsum=0fori=1to100step2sum=sum+inextiprintsum

publicclassdemo{/***@paramargs*/publicstaticvoidmain(String[]args){inta=0,b=0;for(inti=1;i

#include#includevoidmain(){voidss(charstr[]);intsum,i;sum=0,i=1;for(i=1;i

假定5险一金固定是1000元,起征点是3500,程序如下：　　#include　　intmain(intargc,char*argv[])　　{　　floatmoney,temp,total,t;　　

publicclasswages{intmoney;inthour;publicwages(){money=0;hour=0;}publicintweekmoney(){returnmoney*hou

#includevoidmain(){inti,sum=0;for(i=3;i

#includemain(){floata,b,s;printf("pleaceinputa,b:\n");scanf("%f%f\n",&a,&b);s=a+b;printf("%f\n",s);/

#!/bin/bashread-p"Enteranum:"numsum=0foriin$(seq$num)dosum=$((sum+i))doneecho"Sumis$sum"

楼上的好像和题目要求不符#include<stdio.h>#define N 5  //定义N的值void main(){ in

#includeintmain(){intlength,width,perimeter;printf("长：");scanf("%d",&length);printf("宽：");scanf("%d"

#includemain(){inti,m=0;for(i=1;i

//分别计算1到1000之间所有3的倍数之和和所有7的倍数之和intmain(){intsumA=0;//3的倍数的和intsumB=0;//7的倍数的和for(inti=1;i