大概相当于是一个构造题，将(r, g, b)映射为函数值，满足大小关系以及最初两组(c0和c1)rgb的值相等

f(r, g, b) = x*r + y*g + z*b
x + y + z = 1

• c0和c1有大小，无法满足，输出-1
• c0和c1有一个相等，假设r0r1相等，就令x = 1, y = 0, z = 0
• c0和c1有交错，假设是r0 > r1, g0 < g1，令

  p = r0 - r1
  q = g1 - g0
  x = q / (p + q)
  y = p / (p + q)
  

  则

  f(c0) - f(c1) = q / (p + q) * q - p / (p + q) * p = 0
  

看懂了之后自己打了一遍
尤其注意的是公式得到的值是double，最后不能直接取整，要用lround（因此得过一次93.3333的高分）

• lround(double)，将浮点值舍入为最接近的整数

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <cstring>
#include <map>
#include <cmath>
#include <ctime>
#include <cstdlib>
using namespace std;#define N 1005int c[N][3];
double f[3];bool bigger(int x, int y)
{if (c[x][0] > c[y][0] && c[x][1] > c[y][1] && c[x][2] > c[y][2]) return true;return false;
}int main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int n;cin >> n;for (int i = 0; i < n; i++){cin >> c[i][0] >> c[i][1] >> c[i][2];}if (bigger(0, 1) || bigger(1, 0)){cout << -1 << endl;return 0;}f[0] = f[1] = f[2] = 0;int fl = 0;for (int i = 0; i < 3; i++){if (c[0][i] == c[1][i]){f[i] = 1;fl = 1;break;}}if (!fl){int x = 0, y = 0;for (int i = 0; i < 3; i++){if (c[0][i] > c[1][i]) x = i;else y = i;}double p = c[0][x] - c[1][x], q = c[1][y] - c[0][y];f[x] = q / (p + q);f[y] = p / (p + q);}for (int i = 0; i < n; i++){double res = 0;for (int j = 0; j < 3; j++){res += f[j] * c[i][j];}cout << lround(res) << endl;}return 0;
}