HDU 1270 小希的数表 (暴力枚举+数学)

时间：2023-03-10 03:38:57
HDU 1270 小希的数表 (暴力枚举+数学)

题意：...

析：我们可以知道，a1+a2=b1,那么我们可以枚举a1，那么a2就有了，并且a1+a3=b2，所以a3就有了，我们再从把里面的剩下的数两两相加，并从b数组中去掉，

那么剩下的最小的就是a4，然后依次可以求出a5，a6....由于a最大才是5000，并且保证有唯一解，那么找到一个就直接退出。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <unordered_map>

#include <unordered_set>

#define debug() puts("++++");

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const double inf = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 1e4 + 5;

const int mod = 2000;

const int dr[] = {-1, 1, 0, 0};

const int dc[] = {0, 0, 1, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c){

    return r >= 0 && r < n && c >= 0 && c < m;

}

int b[maxn], c[maxn];

int a[maxn];

bool judge(int x){

    a[1] = x;   a[2] = b[1] - a[1];

    int cnt = 2;

    for(int i = 2; i <= m; ++i){

        if(b[i])  a[++cnt] = b[i] - a[1];

        else continue;

        for(int j = 2; j < cnt; ++j){

            bool ok = false;

            for(int k = i+1; k <= m; ++k)  if(a[j] + a[cnt] == b[k]){

                b[k] = 0;

                ok = true;

                break;

            }

            if(!ok)  return ok;

        }

    }

    return true;

}

int main(){

    while(scanf("%d", &n) == 1 && n){

        m = n * (n-1) / 2;

        for(int i = 1; i <= m; ++i)  scanf("%d", c+i);

        for(int i = 1; ; ++i){

            memcpy(b+1, c+1, 4*m);

            if(judge(i))  break;

        }

        for(int i = 1; i <= n; ++i)

            if(i == 1)  printf("%d", a[i]);

            else printf(" %d", a[i]);

        printf("\n");

    }

    return 0;

}

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