給你三個(gè)數(shù)字a，b，c，讓你用1～m的數(shù)字湊出來

結(jié)論：有2個(gè)1和2個(gè)2肯定湊不出來，然后就搜索

#include <bits/stdc++.h> using namespace std; #define rep(i, j, k) for (int i = int(j); i <= int(k); ++ i) #define dwn(i, j, k) for (int i = int(j); i >= int(k); -- i) typedef long long LL; typedef pair<int, int> P; const int N = 3e5 + 7; char col[N]; int dfs(int i, int j, int k, int x) {if (i == 0 && j == 0 && k == 0) return 1;if (i < 0 || j < 0 || k < 0 || x <= 0) return 0;col[x] = 1;if (dfs(i - x, j, k, x - 1)) return 1;col[x] = 2;if (dfs(i, j - x, k, x - 1)) return 1;col[x] = 3;if (dfs(i, j, k - x, x - 1)) return 1;return 0; } int main() {ios::sync_with_stdio(0);LL a, b, c;cin >> a >> b >> c;LL n = a + b + c;LL x = 1;for (x = 1; n > 0; x ++) n -= x; x --; int xx = x;int c1 = (a == 1) + (b == 1) + (c == 1);int c2 = (a == 2) + (b == 2) + (c == 2);if (c1 >= 2 || c2 >= 2 || n != 0) {printf("Impossible\n");return 0;}dfs(a, b, c, x);// cout << a << ' ' << b << ' ' << c << ' ' << x << '\n';auto tran = [&](int x) -> char {if (x == 1) return 'W';if (x == 2) return 'G';return 'B';};rep(i, 1, xx) printf("%c", tran(col[i])); } /* 11 5 5 */

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轉(zhuǎn)載于:https://www.cnblogs.com/tempestT/p/10661076.html

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