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<p>C语言,编写的程序运行结果是相同的,,</p>
<p><img src="http://img.baidu.com/img/iknow/icn_point.gif"> 悬赏分:10 -</p>
<p>解决时间:2010-3-29 12:51</p>
<p>求1-2/3+3/5-4/7..的前N项之和</p>
<p>#include<stdio.h></p>
<p>int main(void)</p>
<p>{</p>
<p>int numerator,denominator,flag,i,n;</p>
<p>double item, sum;</p>
<p>printf("Enter n:");</p>
<p>scanf("%d", &n);</p>
<p>flag = 1;</p>
<p>numerator = 1;</p>
<p>denominator = 1;</p>
<p>sum = 1;</p>
<p>for(i=1;i<=n;i++){</p>
<p>item = flag*numerator / denominator;</p>
<p>sum = sum+item;</p>
<p>flag = -flag;</p>
<p>numerator = numerator+1;</p>
<p>denominator = denominator+2;</p>
<p>}</p>
<p>printf("sum= %f\n",sum);</p>
<p>return 0;</p>
<p>}</p>
<p>运行后的结果都是2.0000</p>
<p>提问者: 微凉ling - 一级</p>
<p>最佳答案</p>
<p>int numerator,denominator,flag,i,n; //你分子分母都是int</p>
<p>int/int =int 导致2/3=0,3/5=0</p>
<p>要把numerator,denominator设为double</p>
<p>而且sum的初始值是0;</p>
<p>0</p>
<p>回答者:</p>
<p>w251988889 - 二级 2010-3-24 18:07</p>
<p>我来评论>></p>
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