double shot

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• pls never say we would have no chance to restore our relationship once again cuz never is a long word and it is a long road that has no turn.and people always say when God closes the door he will open a window for you.So pls give it a shot because we appreciated each other's value and bright points back then.what if you say so to the one you've been head over heels with on Feb 14?good translation!however,i am asking for advice on such an issue "what i
• 输入1个正整数n,计算下式的前n+1项之和(保留4位小数).要求使用嵌套循环.e = 1＋1/1!＋1/2!＋.＋1/n!这个程序哪里出了问题啊#include "stdio.h"int main(void){int i,j,n;int repeat,ri;double e,product;scanf("%d",&repeat);for(ri = 1; ri
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• 10^(-7);++n){P += ((-1)^(n+1))*(1/(2*n-1));}pi = P*4;cout" target="_blank"> 使用C++,求pi的近似值用下面的公式求pi的近似值pi/4 = 1-1/3+1/5-1/7+.直到最后一项的绝对值小于10的-7次方为止如下式我自己编的#include "stdafx.h"#include "iostream"#include "cmath"#include "conio.h"using namespace std;int _tmain(int argc,_TCHAR* argv[]){int n = 1;double P = 0,pi;for(n=1;1/(2*n-1) > 10^(-7);++n){P += ((-1)^(n+1))*(1/(2*n-1));}pi = P*4;cout
• You only get one shot in a lifetim的中文意思、
• Every time i think of you i get a shot right through into a bolt of blue
• 使用C++,求pi的近似值用下面的公式求pi的近似值pi/4 = 1-1/3+1/5-1/7+.直到最后一项的绝对值小于10的-7次方为止如下式我自己编的#include "stdafx.h"#include "iostream"#include "cmath"#include "conio.h"using namespace std;int _tmain(int argc,_TCHAR* argv[]){int n = 1;double P = 0,pi;for(n=1;1/(2*n-1) {P += ((-1)^(n+1))*(1/(2*n-1));}pi = P*4;cout
• pow(10,-6);i++,n+=2)s=s+pow(-1,i)*j/n;pr" target="_blank"> C语言解答:利用公式：π/4=1-1/3+1/5-1/7+……,求π的近似值,直到最后一项的绝对值小于10-6为止利用公式：π/4=1-1/3+1/5-1/7+……,求π的近似值,直到最后一项的绝对值小于10-6为止  下面是我写的,运行显示pi 是4.00000000000,明显不对.又不知道哪里错了.#include#includevoid main(){double s=0,pi,t,d,c;int i,n,j=1;for(i=2,n=1;fabs(1/n)>pow(10,-6);i++,n+=2)s=s+pow(-1,i)*j/n;printf("%f\n",s);printf("pi=%f\n",pi=s*4);}#include#includevoid main(){double s=0,pi,t,d,c;int i,n,j=1;for(i=2,n=1;fabs(1/n)>pow(10,-6);i++,n+=2)s=s+pow(-1,i)*j/n;pr
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