A. Dima and Continuous Line
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解题说明:此题是判断一连串点中依次两个点组成的半圆集合中是否存在相交的情况,可以暴力循环遍历实现。依次判断每个半圆与其他半圆是否相交。
#include <iostream>#include<algorithm>#include<cstdio>#include<cmath>#include<cstring>#include<cstdlib>using namespace std;int main(){ int n,i,x=0,y=0,j; long a[1000];scanf("%d",&n); for(i=0;i<n;i++){cin>>a[i];}for(i=0;i<n;i++){x=y=0;for(j=i+2;j<n;j++) {if ((a[j]>a[i]&&a[j]>a[i+1])||(a[j]<a[i]&&a[j]<a[i+1])){x++;}else{y++;}}if(x&&y){printf("yes\n");return 0;}}printf("no\n");return 0;}
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