fabs(t)>le# include <stdio.h># include <math.h>void main(){ int s; float n,t,pi; t=1; pi=0; s=1; n=1.0; while(fabs(t)>le-6) { pi=pi+t; n=n+2; s=-s; t=s/n; } pi=pi*4; printf("%f",pi); getch();}用WIN -TC编译结果是 错误

fabs(t)>le# include <stdio.h># include <math.h>void main(){ int s; float n,t,pi; t=1; pi=0; s=1; n=1.0; while(fabs(t)>le-6) { pi=pi+t; n=n+2; s=-s; t=s/n; } pi=pi*4; printf(%f,pi); getch();}用WIN -TC编译结果是 错误 #include #include { int p=1;float eps,n=1.0,t=1,pi=0; eps=1*e-5; while(fabs(4*pi-3#include#include{int p=1;float eps,n=1.0,t=1,pi=0;eps=1*e-5;while(fabs(4*pi-3.14159)>=eps){pi+=t;n+=2;p=-p;t=p/n;}printf(pi=%10.6f ,4*pi);}总说declaration syntax e hdu 2039#include#includeint main(){int t;int flag;float a,b,c,max,min;while(scanf(%d,&t)!=EOF){ while(t--){ flag=1;scanf(%f %f %f,&a,&b,&c); if(a+ba+c){max=a+c;}}else{max=b+c;if(max>a+c){max=a+c;}}if(fabs(a-b)>fabs(b-c)){min=fabs(a-b);if(min while (fabs(t) >= 1e-6); 中fabs(t) 什么意思?求～～ C++程序牛顿法解方程,求纠正#include #include #include double root(int n,double x){double t;while(fabs(root(n + 1,x) - root(n,x)) > 1e-6){root(n + 1,x) = root (n,x) - (cos(root(n,x)) - root(n,x)) / (sin((root(n,x))-1);t = (root(n,x);}return #include#include,main(){int s;float n,t,p;t=1.0;p=0;n=1.0;s=1; while(fabs(t)>=1e-6);{p=p+t;n+=2.0s=-s;t=s/n;}p=4*p;pintf(p=%f ,p);}最后怎么输出--------------------Configuration:Cpp1 - Win32 Debug--------------------Compiling...Error spawning C语言 计算圆周率的近似值 填空#include #include void main(){int s;float n,__(1)__;double t;t=1;pi=0;n=1;s=1;while(fabs(t)>=2e-6){__(2)__;n+=2;s=-s;t=s/n;}__(3)__;printf(pi=%.6f ,pi);} 求pi的近似值#include #include void main() { int i=1; double sum=0; double pi=0,t; do{ sum=sum+1/(float)(i*i); i++;pi=sqrt(6*sum); t=pi-3.14159165;}while(fabs(t)>=1e-5);printf(%f,pi);}为何输不出东西来?错在哪里了? 程序如何修改可以用来计算π/4=1-1/3+1/5-1/7······公式求π的近似值,直到最后一项绝对值小于10^(-6#include #include void main(){float s;int i,t;t=1;for(i=1;;i=i+2){s=s+i;i=i+2;t=-t;i=t/i;if(fabs(t) 计算圆周率的算法算法公式是 PI=4*(1-1/3+1/5-1/7+1/9-.)请看下面这个程序#include #include main() {float n,pi,t; int k=1;pi=0,t=1,n=1; while(fabs(t)>1e-6) {t=k/(2*n-1);pi=pi+t; k=-k; n++; } pi=4*pi; printf(%f,pi); } 为什么要把 #include main() {int s,n; float pi=0,t; t=float(s)/n; for(s=1,n=1;fabs(t)>1e-6;s=-s,n+=2) {#includemain(){int s,n;float pi=0,t=1;for(s=1,n=1;fabs(t)>1e-6;s=-s,n+=2){t=float(s)/n;pi=pi+t;}pi=4*pi;printf(%10.6f,pi);} 解决了 自己 可是答案3.14 利用公式求cos x的近似值,精确度为10-6 #include#define PI 3.14159main(){int n,xx;float t,sum,x;printf(请输入弧度x=π/xx对应的分母值:);scanf(%d,&xx);x=PI/xx;t=x;n=0;sum=0;do{sum+=t;t=-t*x*x/(n+1)*(n+2);n+=2;}while(fabs(t)>=1e 在C语言中用Π/4≈1-1/3+1/5-1/+.公式求π的近似值,知道某一项的绝对值小于10-6次方为止#include#includevoid main(){int s;float n,t,pi;t=1;pi=0;n=1.0;s=1;while(fabs(t)>1e-6){pi=pi+t;n=n+2;s=-s;t=s/n;}pi=pi*4;printf(pi=%10.6f C语言中求pi值,语句顺序不同,结果不同,哪位大虾能给我解释下程序是怎么个运行过程#include /*第一种*/main(){double s,n,t,pi;s=1;n=1;pi=0;t=1;while(fabs(t)>1e-6){pi=pi+t;s=-s;n=n+2;t=s/n; /*此语句在最后*/ }pi=4*pi C语言题目：while(fabs(t)=1e-5&&t-10&&s fabs(a) fabs(disc) 编写程序,用如下公式计算圆周率的近似值PI=4-4/3+4/5-4/7+4/9-4/11+.要计算多少项才能得到数值3.14我的程序错在哪里?#include#includeint main(void){int i,j=1,d=1;double PI=0,t;while(fabs(t)>1e-6){t=d*4.0/j;PI+=t;if((int)