可以用一个for循环依次累加就可以解决：
//参考代码#include <stdio.h>
int main()
{
float a=0;
int i,j=1,t;
for(i=1;i<=10;i++){//分母从1到10循环累加
a+=1.0*j/i;//累加和
j*=-1;//控制符号，是加正数还是加负数
}
printf("%f\n",a);
return 0;
}
/*运行结果：
0.645635
*/
using std::cout;
double f(const int &n){
if(n/2)return 1.0/n;
else return -1.0/n;
}
int main()
{
int n(10);
double res(0.0);
while(n)
{
res+=f(n);
--n;
}
cout<<res;
}本回答被提问者采纳
#include <math.h>
void main()
{
float lResult = 1.0;
int i = 2;
while(i<=10)
{
lResult += (1.0/i)*pow(-1.0, i-1);
i++;
}
cout<<"result = "<<lResult<<endl;
}
c++问题:1+1\/2+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8+1\/9+1\/10=?
include <iostream> using namespace std;int main() { double sum = 0.0;for (int i=1;i<=10;i++) { sum += 1.0\/(double)i;} cout << "the sum is " << sum << endl;return 0;}
用C++编写程序函数求表达式1-1\/2+1\/3-1\/4+1\/5-1\/6+1\/7-...+1\/n的值
int k = 1;for(int i = 1; i <= n; i++){ sum += (1.0\/double(i))*k;k = k*(-1);} return sum;}
比较1\/2+1\/3+1\/4+1\/5+1\/6+.+1\/19+1\/20与3的大小,说明理由
即得 1\/2+1\/3+1\/4+1\/5+1\/6+...+1\/19+1\/20<3 1\/2^2+1\/3^2+1\/4^2+...+1\/2002^2与2001\/2002的大小,并说明理由. 裂项(怀疑第一项少抄个1) 1+1\/2^2+1\/3^2+1\/4^2+...+1\/2002^2 > 1\/(1*2)+1\/(2*3)+1\/(3*4)+1\/(4*5)+...+1\/(2002*2003) ...
数列求和 1+1\/2+1\/3+1\/4+1\/5+……1\/n=? 急~
而1+1\/2+1\/3+1\/4+1\/5+...+1\/n (n为无限大)不存在循环节，不可能根据等比数列知识划成两个互质整数相除的形式。所以它终究是无理数。这是有名的调和级数,是高数中的东西。这题目用n!当n->∞，1+1\/2+1\/3+1\/4+1\/5+...+1\/n->∞，是个发散级数 当n很大时，有个近似公式：1...
c++程序设计设s=1+1\/2+1\/3+...+1\/n,求与八最接近的s的值与其对应的n值...
取小者(相等时取前一项)的最后一项的n便是题解。代码如下：include "stdio.h"int main(int argc,char *argv[]){int n;double s;s=n=0;do{s+=1.0\/++n;}while(s<8);if(s-8 > 8-s+1.0\/n)s-=1.0\/n--;printf("s = %f\\tn = %d\\n",s,n);return 0;}运行结果如下：...
C++利用公式pi\/4=1-1\/3+1\/5-1\/7+、、、求pi的近似值,直到最后一位的绝对...
include <iostream> include<iomanip> int main(){ int i = 1;\/\/分母，每次循环+2 int flag = 1;\/\/符号，每次循环变号 double tmp;\/\/存储每次的那一位，tmp = 1.0\/i double sum = 0.0;\/\/存储和，循环结束后 sum = PI\/4 do { tmp = 1.0\/i; \/\/计算每一位 sum = sum +...
用c++求1+1\/2+1\/3+1\/4+1\/n的近似值,要求至少累加到1\/n不大于0.00984为止...
include <stdio.h>int main(){int n; double s=0; n=1; do {s+=1.0\/n; }while(1.0\/n++>0.00984); printf("n=%d\\ns=%.8lf\\n",--n,s); return 0;}
c++编程题:用数组计算序列1\/2.2\/3.3\/5.5\/8. ...的前100项之和_百度知 ...
代码如下：此题主要考察数组的知识。include <stdio.h> include <math.h> int main(int argc, char *argv[]){ int n,i,a=1,b=1,y=1; float sum=0;scanf("%d",&n);for(i=1;i<=n;i++) { Sum+=((float)a\/b)*y;b=a+b; a=b-a; y*=(-1); } printf("%6f"...
用c++,求求各位大佬了!设 s=1+1\/2+1\/3+...+1\/n,求与8最接近的s值及与...
C++代码和运行结果如下：可见与8最接近的s为8.00049，与之对应的n为1673 附C++源码：include <iostream> using namespace std;int main() { double s = 0, last_s;int n = 0;while (s < 8) { n++;s += 1.0 \/ n;last_s = s;} cout << "1+1\/2+1\/3+...+1\/";if (8...
用c++编写1-2+3-4+5-6+7-8+9一直到一百
int main(){ int sum = 0;for(int i=1;i<=100;i++){ if(i%2==0){ sum = sum-i;} else{ sum = sum +i;} } cout<<sum<<endl;return 0;}
