易语言求和
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#includemain(){inti=1,j=1;intsum=0;for(j;j
#includemain(){\x09intm,sum=0;\x09for(m=1;m
有两个问题,一个是f函数逻辑上有问题,第二个主函数调用有问题,sum=sum+f(i)而不是n,修改如下:#includeintf(intn);voidmain(void){intn,sum=0;sc
varf:array[0..1000000]oflongint;x,y,ans,i,max:longint;beginread(x);read(y);f[1]:=1;f[2]:=1;ifx>ythen
递归法写的,变量类型长整数型,长整数型的取值范围为[ -9,223,372,036,854,775,808,9,223,372,036,854,775,807],所以项数设置过高的话,计算结
floatsum=0.0f;floatu,a,b;while(1){scanf("%f%f",&u,&a);b=u+a;sum+=b;printf("这里是每次输入和:%f",b);printf("这
你好! hehe的函数内容改了,你对照原来的程序看看吧,满意请采纳#include<stdio.h>#include<math.h>
#includeintmain(void){\x09inta[10];\x09inti,aver=0;\x09scanf("%d%d%d%d%d%d%d%d%d%d",&a[0],&a[1],&a[2
楼主贴代码,我好给你改啊#include<stdio.h>int main (void){ int sum&nb
#includeusingnamespacestd;intmain(){intnum[5];intsum=0;cin>>a;for(inti=0;i>num[i];if(num[i]
#includeintmain(){\x09inti,sum=0;\x09for(i=1;i
解题思路:利用向量数量积的计算公式来解答。解题过程:解答过程见附件最终答案:略
为了方便,用整数相加举例.#includevoidmain(){inta,b,c,sum;printf("请输入3个数用来相加:\n");scanf("%d%d%d",&a,&b,&c);sum=a+
#include <stdio.h>int main(){int a,b;scanf("%d%d",&a,&b);prin
#include<stdio.h>void main(){\x09int i, n, sum = 0;\x09int
varn,i,x,sum:longint;beginreadln(n);sum:=0;fori:=1tondobeginread(x);inc(sum,x);end;writeln(sum);end.
循环条件里可以有scanf.这样while(scanf("%d",&n)!=EOF)就可以#includeintmain(){intn,a,i,s;while(scanf("%d",&n)!=EOF)
#includevoidmain(){inti,j;inttemp,res1=0,res2=0;for(i=0;i
#include#includeintmain(void){\x05intm;\x05scanf("%d",&m);\x05while(m--){\x05\x05intn,i;\x05\x05doub
解题思路:数列求和解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/readq.ph