int s=1; double n=1,t=1,pi=0; while((fabs(t)) >1e-7) {pi=pi+t; n=n+2; s=-s; t=s/n; } pi=pi*4; co大师们,感激不尽啊 初学者不容易啊
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![int s=1; double n=1,t=1,pi=0; while((fabs(t)) >1e-7) {pi=pi+t; n=n+2; s=-s; t=s/n; } pi=pi*4; co大师们,感激不尽啊 初学者不容易啊](/uploads/image/z/6965800-16-0.jpg?t=int+s%3D1%3B+double+n%3D1%2Ct%3D1%2Cpi%3D0%3B+while%28%28fabs%28t%29%29+%3E1e-7%29+%7Bpi%3Dpi%2Bt%3B+n%3Dn%2B2%3B+s%3D-s%3B+t%3Ds%2Fn%3B+%7D+pi%3Dpi%2A4%3B+co%E5%A4%A7%E5%B8%88%E4%BB%AC%2C%E6%84%9F%E6%BF%80%E4%B8%8D%E5%B0%BD%E5%95%8A+%E5%88%9D%E5%AD%A6%E8%80%85%E4%B8%8D%E5%AE%B9%E6%98%93%E5%95%8A)
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int s=1; double n=1,t=1,pi=0; while((fabs(t)) >1e-7) {pi=pi+t; n=n+2; s=-s; t=s/n; } pi=pi*4; co大师们,感激不尽啊 初学者不容易啊
int s=1; double n=1,t=1,pi=0; while((fabs(t)) >1e-7) {pi=pi+t; n=n+2; s=-s; t=s/n; } pi=pi*4; co
大师们,感激不尽啊 初学者不容易啊
int s=1; double n=1,t=1,pi=0; while((fabs(t)) >1e-7) {pi=pi+t; n=n+2; s=-s; t=s/n; } pi=pi*4; co大师们,感激不尽啊 初学者不容易啊
这是典型的求和表示,循环中做的是:
pi=1-1/3+1/5-1/7+……
循环后pi=pi/4,这求出的就是圆周率的约值.