CSU 1165: Nearest Numbers
【摘要】
题目：
Description
Given three integer sequences which have been sorted in ascending order,pick one integer from each sequence and we can get three integers.we wa...
题目：
Description
Given three integer sequences which have been sorted in ascending order,pick one integer from each sequence and we can get three integers.we want these three integers to be nearest to each other.Assuming these three integers as a,b,c, they are nearest to each other means d=(a-b)2+(b-c)2+(c-a)2 is the smallest.
Input
There are many test cases. For each test case,the first line are three integers la,lb,lc, they let you know the length of each sequence. The following three lines separately have la,lb,lc integers,give you the integers in each sequence. 1<=la,lb,lc<=10^6 and All integers in the sequences are in the range of [-10^9,10^9].
AC代码：
#include<iostream>
#include<algorithm>
using namespace std;
long long x[3][100001];
long long f(int a, int b, int c)
{
long long r = 1;
r = (r << 63) - 1;
for (int i = 1, j = 1, k = 1; i <= x[a][0]; i++)
{
while (j < x[b][0] && x[a][i] >= x[b][j+1])j++;
while (k < x[c][0] && x[a][i] > x[c][k])k++;
r = min(r,(x[a][i] - x[b][j])*(x[a][i] - x[b][j]) + (x[b][j] - x[c][k])*(x[b][j] - x[c][k]) + (x[a][i] - x[c][k])*(x[a][i] - x[c][k]));
}
return r;
}
int main()
{
while (cin >> x[0][0] >> x[1][0] >> x[2][0])
{
for (int i = 0; i < 3; i++)for (int j = 1; j <= x[i][0]; j++)cin >> x[i][j];
long long r1 = min(f(0, 1, 2), f(0, 2, 1));
long long r2 = min(f(1, 0, 2), f(1, 2, 0));
long long r3 = min(f(2, 0, 1), f(2, 1, 0));
cout << min(min(r1, r2), r3) << endl;
}
return 0;
}
文章来源: blog.csdn.net，作者：csuzhucong，版权归原作者所有，如需转载，请联系作者。
原文链接：blog.csdn.net/nameofcsdn/article/details/79726353
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