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POJ2718-Smallest Difference-窮竭搜索

2019-11-08 18:28:29
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原題鏈接 Smallest Difference Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 10032 Accepted: 2752 Description

Given a number of distinct decimal digits, you can form one integer by choosing a non-empty subset of these digits and writing them in some order. The remaining digits can be written down in some order to form a second integer. Unless the resulting integer is 0, the integer may not start with the digit 0.

For example, if you are given the digits 0, 1, 2, 4, 6 and 7, you can write the pair of integers 10 and 2467. Of course, there are many ways to form such pairs of integers: 210 and 764, 204 and 176, etc. The absolute value of the difference between the integers in the last pair is 28, and it turns out that no other pair formed by the rules above can achieve a smaller difference. Input

The first line of input contains the number of cases to follow. For each case, there is one line of input containing at least two but no more than 10 decimal digits. (The decimal digits are 0, 1, …, 9.) No digit appears more than once in one line of the input. The digits will appear in increasing order, separated by exactly one blank space. Output

For each test case, write on a single line the smallest absolute difference of two integers that can be written from the given digits as described by the rules above. Sample Input

1 0 1 2 4 6 7 Sample Output

28 Source 題意:這些單個的數字組成的兩個數的最小差 思路:通過字典序排列排好后一分成兩個數字,求此最小差 Rocky Mountain 2005

#include <cstdio>#include <string>#include <cstring>#include <sstream>#include <iostream>#include <algorithm>using namespace std;const int INF = 0x3f3f3f3f;const int maxn = 11;int a[maxn];//讀取一行中的數組并且存儲int read_list(){ string line; getline(cin,line); stringstream ss(line); int t=0,x; while(ss >> x) a[t++]=x; return t;}int abs(int x){ return x > 0 ? x : -x;}int main(){ int K; cin >> K; string line; getline(cin,line);//使用getline總是需要先預讀一個空行 while(K--){ int res=INF; int n=read_list(); if(n==2){ cout << abs(a[0]-a[1]) << endl; continue; } int mid=(n+1)/2;//如果mid=n/2那么左邊一定不會是一個0,而右邊也最少兩個元素無法生成一個單獨的0 while(a[0]==0) next_permutation(a,a+n);//只要新生成的排列的首字符是0就繼續進行排列 do{ if(a[mid]==0 && n-mid == 1){//要么有且僅有一個0 int l=a[0]; for(int i=1;i<mid;i++) l=l*10 + a[i]; res = min(res,l); } if(a[mid]){//要么首字符不為0 int l=a[0],r=a[mid]; for(int i=1;i<mid;i++) l = l*10 + a[i]; for(int i=mid+1;i<n;i++) r = r*10 + a[i]; res = min(res,abs(l-r)); } }while(next_permutation(a,a+n)); cout << res << endl; } return 0;}
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