网站后台 灰色,经营网站需要什么资质,建设网站论文范文,会计信息网站建设的意思传送门 文章目录题意#xff1a;思路#xff1a;题意#xff1a;
给你三个数组a,b,ca,b,ca,b,c#xff0c;让你从每个数组中选择一个数x,y,zx,y,zx,y,z#xff0c;使得(x−y)2(x−z)2(y−z)2(x-y)^2(x-z)^2(y-z)^2(x−y)2(x−z)2(y−z)2最小#xff0c;求这个最小值。
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文章目录题意思路题意
给你三个数组a,b,ca,b,ca,b,c让你从每个数组中选择一个数x,y,zx,y,zx,y,z使得(x−y)2(x−z)2(y−z)2(x-y)^2(x-z)^2(y-z)^2(x−y)2(x−z)2(y−z)2最小求这个最小值。
思路
由于答案一定是x≤y≤zx\le y\le zx≤y≤z的形式(当然这里的x,y,zx,y,zx,y,z与题意的x,y,zx,y,zx,y,z不同)所以我们就可以跑一个3!3!3!排列让他们分别为x,y,zx,y,zx,y,z让后枚举其中一个二分另外俩就行了。 具体实现可以写个函数就比较方便了。 我这里写的是枚举xxx让后分两种即找yyy和找zzz如果找yyy的话假设现在已经有x≤yx\le yx≤y了那么找的zzz根据平方的性质最好是x(y−x)/2x(y-x)/2x(y−x)/2附近的位置二分找即可对于zzz同理。 总结这个题瞎搞都能过。
// Problem: D. Xenia and Colorful Gems
// Contest: Codeforces - Codeforces Round #635 (Div. 2)
// URL: https://codeforces.com/contest/1337/problem/D
// Memory Limit: 256 MB
// Time Limit: 3000 ms
//
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize(Ofast,no-stack-protector,unroll-loops,fast-math)
//#pragma GCC target(sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tunenative)
//#pragma GCC optimize(2)
#includecstdio
#includeiostream
#includestring
#includecstring
#includemap
#includecmath
#includecctype
#includevector
#includeset
#includequeue
#includealgorithm
#includesstream
#includectime
#includecstdlib
#define X first
#define Y second
#define L (u1)
#define R (u1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].ltr[u].r1)
#define Len(u) (tr[u].r-tr[u].l1)
#define random(a,b) ((a)rand()%((b)-(a)1))
#define db puts(---)
using namespace std;//void rd_cre() { freopen(d://dp//data.txt,w,stdout); srand(time(NULL)); }
//void rd_ac() { freopen(d://dp//data.txt,r,stdin); freopen(d://dp//AC.txt,w,stdout); }
//void rd_wa() { freopen(d://dp//data.txt,r,stdin); freopen(d://dp//WA.txt,w,stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pairint,int PII;const int N1000010,mod1e97,INF0x3f3f3f3f;
const LL inf0x3f3f3f3f3f3f3f3f;
const double eps1e-6;int n1,n2,n3;
LL a[N],b[N],c[N];LL get(int i,int j,int k) {//couti j kendl;return (a[i]-b[j])*(a[i]-b[j])(a[i]-c[k])*(a[i]-c[k])(b[j]-c[k])*(b[j]-c[k]);
}LL solve() {LL ansinf;for(int i1;in1;i) {int val1a[i];int pos1lower_bound(b1,b1n2,val1)-b;if(pos1n2) {int pos2lower_bound(c1,c1n3,a[i](b[pos1]-a[i])/2)-c;if(pos2n3) ansmin(ans,get(i,pos1,pos2)); pos2--;if(pos21) ansmin(ans,get(i,pos1,pos2));}pos1;if(pos1n2) {int pos2lower_bound(c1,c1n3,a[i](b[pos1]-a[i])/2)-c;if(pos2n3) ansmin(ans,get(i,pos1,pos2)); pos2--;if(pos21) ansmin(ans,get(i,pos1,pos2));}pos1-2;if(pos11) {int pos2lower_bound(c1,c1n3,b[pos1](a[i]-b[pos1])/2)-c;if(pos2n3) ansmin(ans,get(i,pos1,pos2)); pos2--;if(pos21) ansmin(ans,get(i,pos1,pos2));}pos1lower_bound(c1,c1n3,val1)-c;if(pos1n3) {int pos2lower_bound(b1,b1n2,a[i](c[pos1]-a[i])/2)-b;if(pos2n2) ansmin(ans,get(i,pos2,pos1)); pos2--;if(pos21) ansmin(ans,get(i,pos2,pos1));}pos1;if(pos1n3) {int pos2lower_bound(b1,b1n2,a[i](c[pos1]-a[i])/2)-b;if(pos2n2) ansmin(ans,get(i,pos2,pos1)); pos2--;if(pos21) ansmin(ans,get(i,pos2,pos1));}pos1-2;//coutpos1 c[pos1] a[i]endl;if(pos11) {int pos2lower_bound(b1,b1n2,c[pos1](a[i]-c[pos1])/2)-b;if(pos2n2) ansmin(ans,get(i,pos2,pos1)); pos2--;if(pos21) ansmin(ans,get(i,pos2,pos1));}}return ans;
}int main()
{
// ios::sync_with_stdio(false);
// cin.tie(0);int _; scanf(%d,_);while(_--) {scanf(%d%d%d,n1,n2,n3);for(int i1;in1;i) scanf(%lld,a[i]);for(int i1;in2;i) scanf(%lld,b[i]);for(int i1;in3;i) scanf(%lld,c[i]);sort(a1,a1n1); sort(b1,b1n2); sort(c1,c1n3);n1unique(a1,a1n1)-a-1;n2unique(b1,b1n2)-b-1;n3unique(c1,c1n3)-c-1;printf(%lld\n,solve());}return 0;
}
/**/
