标签:for tac double etl link char main 1.0 ack
题意:将n个数字分成两组,两组分别组成一个数字,问两个数字的最小差值。要求,当组内数字个数多于1个时,组成的数字不允许有前导0。(2<=n<=10,每个数字范围是0~9)
分析:
1、枚举n个数字的全排列。
2、当两组数字个数相同或只差1时组成的两个数字才可能出现最小差值。
3、0~cnt/2 - 1为前半组数字,cnt/2~cnt-1为后半组数字。
4、注意getchar()的位置。
#pragma comment(linker, "/STACK:102400000, 102400000") #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 Min(a, b) ((a < b) ? a : b) #define Max(a, b) ((a < b) ? b : a) 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 = 10 + 10; const int MAXT = 10000 + 10; using namespace std; int a[MAXN]; string s; int cnt; bool judge(){ if(a[0] == 0 && cnt / 2 - 0 > 1) return false;//前半组数字个数大于1 if(a[cnt / 2] == 0 && cnt - cnt / 2 > 1) return false;//后半组数字个数大于1 return true; } int main(){ int T; scanf("%d", &T); getchar(); while(T--){ getline(cin, s); stringstream ss(s); int x; cnt = 0; while(ss >> x){ a[cnt++] = x; } sort(a, a + cnt); int ans = INT_INF; do{ int sum1 = 0, sum2 = 0; if(!judge()) continue; for(int i = 0; i < cnt / 2; ++i){ sum1 = sum1 * 10 + a[i]; } for(int i = cnt / 2; i < cnt; ++i){ sum2 = sum2 * 10 + a[i]; } ans = Min(ans, abs(sum1 - sum2)); }while(next_permutation(a, a + cnt)); printf("%d\n", ans); } return 0; }
POJ - 2718 Smallest Difference(全排列)
标签:for tac double etl link char main 1.0 ack
原文地址:http://www.cnblogs.com/tyty-Somnuspoppy/p/6512346.html