uva 10790 How Many Points of Intersection?(数学)
2017-08-10 14:04
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题意:
给定上下两条线分别有多少个点 ,之后求直线相交之后的交点数目,注意线上面的交点是不算的。
思路:
找规律,先画好上面的a个点,
画下面第一个点:它与上面所有点连线后没有交点。
画下面第二个点:它与a1连线后交点为a-1,与a2连线交点为a-2,........ 与最后一个点连线交点为0;
画下面第三给点:它与a1连线后交点为2*(a-1),与a2连线交点为1(a-2)......与最后一个交点连线交点为0;
以此类推。
注意:上下两条线上面的点的排序是没有等间距的,等间距会导致交点变少。
#include<iostream>
#include<cstdio>
using namespace std;
int main (){
long long m,n;
int co=1;
while(scanf("%lld %lld",&m,&n)&&m&&n){
printf("Case %d: ",co++);
printf("%lld\n",m*(m-1)*n*(n-1)/4);
}
return 0;
}
给定上下两条线分别有多少个点 ,之后求直线相交之后的交点数目,注意线上面的交点是不算的。
思路:
找规律,先画好上面的a个点,
画下面第一个点:它与上面所有点连线后没有交点。
画下面第二个点:它与a1连线后交点为a-1,与a2连线交点为a-2,........ 与最后一个点连线交点为0;
画下面第三给点:它与a1连线后交点为2*(a-1),与a2连线交点为1(a-2)......与最后一个交点连线交点为0;
以此类推。
注意:上下两条线上面的点的排序是没有等间距的,等间距会导致交点变少。
#include<iostream>
#include<cstdio>
using namespace std;
int main (){
long long m,n;
int co=1;
while(scanf("%lld %lld",&m,&n)&&m&&n){
printf("Case %d: ",co++);
printf("%lld\n",m*(m-1)*n*(n-1)/4);
}
return 0;
}
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