按不同精度求圆周率.这个C++程序出了什么问题,题目是PI/4=1-1/3+1/5-1/7...求PI的代码如下#includeusing namespace std;int MI(int j){int y=1;for(int i=1;i>m;for(j=1;j
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![按不同精度求圆周率.这个C++程序出了什么问题,题目是PI/4=1-1/3+1/5-1/7...求PI的代码如下#includeusing namespace std;int MI(int j){int y=1;for(int i=1;i>m;for(j=1;j](/uploads/image/z/8661752-8-2.jpg?t=%E6%8C%89%E4%B8%8D%E5%90%8C%E7%B2%BE%E5%BA%A6%E6%B1%82%E5%9C%86%E5%91%A8%E7%8E%87.%E8%BF%99%E4%B8%AAC%2B%2B%E7%A8%8B%E5%BA%8F%E5%87%BA%E4%BA%86%E4%BB%80%E4%B9%88%E9%97%AE%E9%A2%98%2C%E9%A2%98%E7%9B%AE%E6%98%AFPI%2F4%3D1-1%2F3%2B1%2F5-1%2F7...%E6%B1%82PI%E7%9A%84%E4%BB%A3%E7%A0%81%E5%A6%82%E4%B8%8B%23includeusing+namespace+std%3Bint+MI%28int+j%29%7Bint+y%3D1%3Bfor%28int+i%3D1%3Bi%3Em%3Bfor%28j%3D1%3Bj)
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