HDU - 6168 Numbers
题意:已知序列a和b的混合序列,序列b是由序列a中满足i < j的ai+aj所组成,要求升序输出序列a。
分析:
1、将混合序列c排序,则第一个元素一定为序列a中的最小值,即输出序列中的a[1],同理c[2] = a[2]。删去c[1],c[2]。
2、由此可知a[1] + a[2],并将其从序列c中删去,此时序列c中的最小值一定为a[3]。删去a[3]。
3、同理,可知a[1] + a[3]和a[2] + a[3]删去,此时序列c中的最小值一定为a[4]。
#include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include<sstream> #include<iterator> #include<algorithm> #include<string> #include<vector> #include<set> #include<map> #include<stack> #include<deque> #include<queue> #include<list> #define lowbit(x) (x & (-x)) const double eps = 1e-8; inline int dcmp(double a, double b){ if(fabs(a - b) < eps) return 0; return a > b ? 1 : -1; } typedef long long LL; typedef unsigned long long ULL; const int INT_INF = 0x3f3f3f3f; const int INT_M_INF = 0x7f7f7f7f; const LL LL_INF = 0x3f3f3f3f3f3f3f3f; const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f; const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1}; const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1}; const int MOD = 1e9 + 7; const double pi = acos(-1.0); const int MAXN = 125250 + 10; const int MAXT = 100000 + 10; using namespace std; int a[510]; int c[MAXN]; map<int, int> mp; int main(){ int m; while(scanf("%d", &m) == 1){ mp.clear(); for(int i = 1; i <= m; ++i){ scanf("%d", &c[i]); ++mp[c[i]]; } if(m == 0){ printf("0\n"); continue; } if(m == 1){ printf("1\n%d\n", c[1]); continue; } sort(c + 1, c + m + 1); int n = (int)(sqrt(double(1 + 8 * m)) - 1) / 2; a[1] = c[1]; a[2] = c[2]; --mp[c[1]]; --mp[c[2]]; --mp[a[1] + a[2]]; int k = 3; int cnt = 3; while(cnt <= n){ if(mp[c[k]] == 0){ ++k; continue; } a[cnt] = c[k]; --mp[c[k]]; for(int i = 1; i <= cnt - 1; ++i){ --mp[a[i] + a[cnt]]; } ++cnt; ++k; } printf("%d\n", n); for(int i = 1; i <= n; ++i){ if(i != 1) printf(" "); printf("%d", a[i]); } printf("\n"); } return 0; }