C语言，求1+3+5+.......+n的编程 用C语言编程求1-3+5-7+9-11+...-99+101...
\u600e\u6837\u5229\u7528C\u8bed\u8a00\u7f16\u7a0b\u6c421!+3!+5!+\u2026\u2026+N!\u7684\u503c\uff1f#include
int main(){
int n,i;
long a,sum = 1;
printf("\u8bf7\u8f93\u5165\u4e00\u4e2a\u5947\u6570:\n");
scanf("%ld",&n);
for(i = 3,a = 1;i <= n;i += 2)
{
a *= i * (i - 1);
sum += a;
}
printf("%ld\n",sum);
return 0;
}
\u672c\u7a0b\u5e8f\u5df2\u7ecf\u5728VC\u4e0a\u8fd0\u884c\u901a\u8fc7\uff0c\u5728N\u4e0d\u662f\u5f88\u5927\u7684\u60c5\u51b5\u4e0b\u53ef\u4ee5\uff0c\u592a\u5927\u4e86\u5c31\u5c5e\u4e8e\u5927\u6570\u95ee\u9898\u4e86\uff0c\u8981\u6539
#include"stdio.h"
main()
{ int i,j,sum=0;
for(i=1,j=1;i<=101;i=i+2)
{sum+=i*j;
j=j*(-1);
}
printf(" sum=%d ",sum);
}
\u6211\u8fd9\u91cc\u6ca1\u6709\u7f16\u8bd1\u5668\uff0c\u8fd8\u6ca1\u6709\u8c03\u8bd5\u3002 \u4e0d\u8fc7\uff0c\u5e94\u8be5\u53ef\u4ee5\u5728visual c++\u4e2d\u8fd0\u884c\u3002
#include<stdio.h>
int main()
{
int n;
do{
scanf("%d",&n);//从键盘输入一个奇数给n
}while(n%2==0);
for(int sum=0,i=1;i<=n;i+=2)
sum+=i;
printf("sum=%d\n",sum);
return 0;
}
有技术含量的，递归求解
#include<stdio.h>
int a(int n)
{
if (n==1)
return 1;
else
return a(n-1)+n;
}
void main()
{
int n;
printf("输入一个数N：");
scanf("%d",&n);
printf("1+2+...n= %d\n",a(n));
}
i=m*i;这句把i改了，改成
#include<stdio.h>
void main()
{
int m=1,i,n,sum=0;
printf("ÇëÊäÈëÒ»¸öNµÄÖµ£º");
scanf("%d",&n);
for(i=1;i<=n+1;i++)
{
sum+=i*m;
m=-m;
}
printf("sum=%d\n",sum);
}
就好了
#include<stdio.h>
int main()
{
int n,i,sum=0;
printf("请输入一个整数：\n");
do{
scanf("%d",&n);//从键盘输入一个奇数给n
}while(n%2==0);
for(i=1;i<=n;i=i+2)
sum=sum+i;
printf("sum=&d\n");
return 0;
}
这个算是简明易懂的了，有不懂的可以问我哈
int sum = (a[0]+(n-1)*d)*n/2; //d为公差,d=2 ,n为数的总个数
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