...{ count++; m=m&(m-1); } printf("count=%d\n",count);

for (index = 0; index < 5; index++)array3[index] = array1[index];for (index = 0; index < 5; index++)array3[index+5] = array2[index];sort(array3, 10);printf("合并、排序后的数组为：\n");for (index = 0; index < 10; index++)printf("%d ", array3[index]);p...

while((n >= 2) && (count < 10)) { for(i = 2; i * i <= n; ++i) if(n % i == 0) break; if (i * i > n) { sum += n; ++count; } --n; } return sum;}int main(void){ printf("%d\n", fun(100)); return 0;} 已赞过 已踩过< 你对这个回答的评价是? 评论...

val[sp++] = f;else printf("error: stack full, can't push %g\n", f);} /* pop:pop and return top value from stack */ double pop(void) { if (sp > 0)return val[--sp];else { puts("error: stack empty\n");return 0.0;} } / Definition of getOp()./ include <...

insertcount++; } } else { errorcount++; } } Response.Write((insertcount + "条数据导入成功!" + updatecount + "条数据重复!" + errorcount + "条数据部分信息为空没有导入!")); } catch(Exception ex) { } } 本回答由电脑网络分类达人 赵丽丽推荐 举报| 答案纠错 | 评论...

<?php str = '1-1-2';arr = explode( '-', $str );arr[0] = $arr[0].'单元';arr[1] = $arr[1].'楼';arr[2] = $arr[2].'号';str = $arr[0].$arr[1].$arr[2];echo $str; //输出 1单元1楼2号 ?>

static extern bool QueryPerformanceFrequency(ref long count);static void SubTest(){ long count = 0;long count1 = 0;long freq = 0;double result = 0;QueryPerformanceFrequency(ref freq);QueryPerformanceCounter(ref count);//耗时巨大的代码 QueryPerformanceCounter(ref count1);count = count...

void countValue(){ int i;for(i=100;i<=999&&3*i/2<=9999;i++)if(3*i%2==0&&3*i/2>1000&&3*i/2<=9999){ cnt++;sum+=i+3*i/2;} 这里是不正确的，缺少判断要 满足SIX+SIX+SIX=NINE+NINE,i相同，n相同 你的做法是可以的，可以去掉外层的if判断了(显得多余了)!内层的if...