poj 1658 Eva's Problem
2014-07-16 19:06
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确实是非常水的题,在这里留个痕迹
AC的代码:
AC的代码:
#include<stdio.h> int main() { int n; scanf("%d",&n); int a[6],i; int gap; int result; while(n--) { for (i=1;i<=4;i++) scanf("%d",&a[i]); //等差更容易判断,如果不是等差就一定是等比 gap=a[2]-a[1]; if (gap==a[3]-a[2]) { //是等差数列 result=a[4]+gap; } else { //等比数列 result=a[4]*a[2]/a[1]; } for(i=1;i<=4;i++) printf("%d ",a[i]); printf("%d\n",result); } return 0; }
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