1重for循环结构编程,计算39到100之间的奇数的和s

importjava.util.Scanner;publicclassTest2{publicstaticvoidmain(String[]args){System.out.println("下面将计

while语句：#includeinti=0,sum=0;main(){while(i

#includevoidmain(void){intn=0;inti=0;floats=0;floatfTemp=0.0f;scanf("%d",&n);for(i=1;i

publicclassTest{publicstaticvoidmain(String[]args){doublex=0.0;//当前壶中酒for(inti=0;i

clears=0fori=1to10m=1forj=1toim=m*jendfors=s+mendfor?"S=1!+2!+3!+…10!=",s

C语言的.#include"stdio.h"main(){floatn=0;floatsum=0;for(;n

#includeintmain(){intx,i,n;doubley;printf("Inputainteger:");scanf("%d",&x);n=1;y=0.0;for(i=1;i

按照你的说法,是算从0变到1的次数,而不单纯是变化（0变1或者1变0）,用不着什么for,if,直接length(find(diff(a)==1))就可以了如果非要用for,ifs=0;forn=1:

你好!PrivateSubForm_Click()DimiAsIntegerDimsAsIntegerFori=1To100s=s+iNextiPrintsEndSu

functionans=FOR(n)sum=0;fori=1:nsum=sum+i^i;i=i+1;endans=sum;functionans=WHILE(n)sum=0;i=1;whilei

#includeusingnamespacestd;intf(intn){ints=0,t=1;for(inti=1;i

第一题：programceshi1;varm,n,r,k:longint;beginreadln(m,n);k:=m*n;r:=mmodn;whiler0dobeginm:=n;n:=r;r:=mmo

#include//whileintexp1(inttemp1){inti=1;intterm=1;while(i

intn=100;intsum=0;for(inti=0;i

sum=0;j=0;fori=1to30step2j=j+1;if(j%2==0)s=-1/i;sum=sum+s;elses=1/isum=sum+s;endifendfo

dimn,s,i,pdon=val(inputbox("大于0的正整数"))loopuntiln>0andn=int(n)s=1p=-1fori=1tons=s+p*2*n/(2*n+1)p=p*-1

#includeintmain(){inta,b,n;scanf("%d%d%d",&a,&b,&n);intcnt=0,f=0;for(inti=a;i再问：没有学过【10】这个，用最简单的循环语句

#includemain(){inta[100],i=0,sum=0;do{scanf("%d",&a[i]);i++;}while(a[i-1]0)//求奇数和{if(a[i

核心代码如下:intsum=0;for(inti=1;i

ivateSubCommand17_Click()DimiAsLong,kAsLong,jAsLong,oAsLongk=Val(Text1.Text)‘要计算的数j=0o=0Fori=1TokIfi