Codeforces 1501C （数学+暴力）

题意描述

给定\(n\)个数(\(n≤200000\))，求是否存在一对(\(x,y\))和(\(z,w\))，使得\(a_{x}+a_{y}=a_{z}+a_{w}\)。

思路

由于题目中每个数字的范围位于[\(1,2.5*10^6\)],所以任意两个数的和都在[\(2,5*10^6\)]区间内。
由鸽巢原理得，枚举\(5*10^6\)对\((i,j)\)后，一定存在一对\((x,y)\)和\((z,w)\)具有相同的\(sum\)。
所以暴力枚举即可，复杂度\(O(min(5*10^6,n^2))\)。

AC代码

#include "iostream"
#include "cstring"
#include "string"
#include "vector"
#include "cmath"
#include "algorithm"
#include "map"
#include "set"
#include "queue"
#include "stack"
#include "cassert"
#include "unordered_map"
#include "sstream"
#include "cstdio"

using namespace std;

#define fi first
#define se second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) a.begin(),a.end()
#define rep(x,l,u) for(ll x=l;x=u;x--)
#define sz(x) x.size()
#define IOS ios::sync_with_stdio(false);cin.tie(nullptr);
#define seteps(N) setprecision(N)
#define uni(x) sort(all(x)), x.erase(unique(all(x)), x.end())
#define lson (ind<<1)
#define rson (ind<<1|1) 
#define endl '\n'
#define dbg(x) cerr << #x " = " << (x) << endl
#define mp make_pair
//#define LOCAL

typedef long long ll;
typedef unsigned long long ull;
typedef __int128 lll;
typedef pair PII;
typedef pair PCC;
typedef pair PDD;
typedef pair PLL;
typedef pair PIII;


struct Scanner {
 
    bool hasNext = 1;
    bool hasRead = 1;
 
    int nextInt() {
        hasRead = 0;
        int res = 0;
        char flag = 1, ch = getchar();
        while(ch != EOF && !isdigit(ch)) {
            hasRead = 1;
            flag = (ch == '-') ? -flag : flag;
            ch = getchar();
        }
        while(ch != EOF && isdigit(ch)) {
            hasRead = 1;
            res = res * 10 + (ch - '0');
            ch = getchar();
        }
        if(ch == EOF)
            hasNext = 0;
        return res * flag;
    }
 
    ll nextLL() {
        hasRead = 0;
        ll res = 0;
        char flag = 1, ch = getchar();
        while(ch != EOF && !isdigit(ch)) {
            hasRead = 1;
            flag = (ch == '-') ? -flag : flag;
            ch = getchar();
        }
        while(ch != EOF && isdigit(ch)) {
            hasRead = 1;
            res = res * 10 + (ch - '0');
            ch = getchar();
        }
        if(ch == EOF)
            hasNext = 0;
        return res * flag;
    }
 
    char nextChar() {
        hasRead = 0;
        char ch = getchar();
        while(ch != EOF && isspace(ch)) {
            hasRead = 1;
            ch = getchar();
        }
        if(ch == EOF)
            hasNext = 0;
        return ch;
    }
 
    int nextString(char *str) {
        hasRead = 0;
        int len = 0;
        char ch = getchar();
        while(ch != EOF && isspace(ch)) {
            hasRead = 1;
            ch = getchar();
        }
        while(ch != EOF && !isspace(ch)) {
            hasRead = 1;
            str[++len] = ch;
            ch = getchar();
        }
        str[len + 1] = 0;
        if(ch == EOF)
            hasNext = 0;
        return len;
    }
 
} sc;
 
ll rd() {
    ll x = sc.nextLL();
    return x;
}
 
void rd(int &x) {
    x = sc.nextInt();
}
 
void rd(ll &x) {
    x = sc.nextLL();
}
 
void rd(char &x) {
    x = sc.nextChar();
}
 
void rd(char* x) {
    sc.nextString(x);
}
 
template
void rd(pair &x) {
    rd(x.first);
    rd(x.second);
}
 
template
void rd(T *x, int n) {
    for(int i = 1; i <= n; ++i)
        rd(x[i]);
}

template
void rd(vector &x,int n){
    for(int i = 1; i <= n; ++i)
        rd(x[i]);
}
 
void printInt(int x) {
    if(x < 0) {
        putchar('-');
        x = -x;
    }
    if(x >= 10)
        printInt(x / 10);
    putchar('0' + x % 10);
}
 
void printLL(ll x) {
    if(x < 0) {
        putchar('-');
        x = -x;
    }
    if(x >= 10)
        printLL(x / 10);
    putchar('0' + x % 10);
}
 
void pr(int x, char ch = '\n') {
    printInt(x);
    putchar(ch);
}
 
void pr(ll x, char ch = '\n') {
    printLL(x);
    putchar(ch);
}

template
void pr(pair x, char ch = '\n') {
#ifdef LOCAL
    putchar('<');   
    pr(x.first, ' ');
    pr(x.second, '>');
    putchar(ch);
    return;
#endif //LOCAL
    pr(x.first, ' ');
    pr(x.second, ch);
}
template
void pr(T *x, int n) {
    for(int i = 1; i <= n; ++i)
        pr(x[i], " \n"[i == n]);
}
 
template
void pr(vector &x) {
    int n = x.size();
    for(int i = 1; i <= n - 1; ++i)
        pr(x[i], " \n"[i == n - 1]);
}

const int N=5*1e6+10;
const int M=1<<12;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
const lll oone=1;
const double eps=1e-6;
const double pi=acos(-1);

int n,a[N];
PII b[N];
struct Solver
{
    void InitOnce(){

    }

    void Read(){
        rd(n);
        rd(a,n);
    }

    void Solve(){
        rep(i,1,n+1){
            rep(j,i+1,n+1){
                int sum=a[i]+a[j];
                if(b[sum].fi){
                    if(i!=b[sum].fi && j!=b[sum].se && i!=b[sum].se && j!=b[sum].fi){
                        puts("YES");
                        pr(i,' ');pr(j,' ');pr(b[sum].fi,' ');pr(b[sum].se);
                        return;
                    }
                }else{
                    b[sum]={i,j};
                }
            }
        }
        puts("NO");
    }
}solver;
int main(){
#ifdef LOCAL
    freopen("data.in","r",stdin);
#endif  //LOCAL
    solver.InitOnce();
    int t=1;
    //t=sc.nextInt();
    //t=INF;
    while(t--){
        solver.Read();
        if(!sc.hasRead) break;
        solver.Solve();
        if(t>=1) puts("");
        if(!sc.hasNext) break;
    }
}