POJ 2718 Smallest Difference 未AC 《挑战程序设计竞赛》

题目：POJ 2718

思路：

分为奇数和偶数两种情况进行处理，输入个数为奇数的时候，无须穷举，最小差是一定的，如0 1 2 3 4，最小差为102 - 43。

输入个数为偶数的时候，用next_permutation穷举。

没有AC……

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <iostream>
 4 #include <stdio.h>
 5 #include <string.h>
 6
 7 using namespace std;
 8
 9 int n;
10 int input[28];
11 char num[128];
12
13 long long pow10(int n)
14 {
15     long long ret = 1;
16     for (int i = 0 ;i < n; i++)
17     {
18         ret *= 10;
19     }
20     return ret;
21 }
22
23 int getNumber(int * addr, int count, bool isMin) {
24     int res = 0;
25     if (isMin) {
26         if (addr[0] == 0) {
27             swap(addr[0], addr[1]);
28         }
29         for (int i = count - 1; i >= 0; i--) {
30             res += *(addr + (count - 1 - i)) * pow10(i);
31         }
32     }
33     else {
34         for (int i = 0; i < count; i++) {
35             res += *(addr + i) * pow10(i);
36         }
37     }
38     return res;
39 }
40
41 int main() {
42     cin >> n;
43     cin.ignore();
44     int lc;    //left count
45     int rc;   //right count
46
47     int f;
48     int diff;
49     for (int i = 0; i < n; i++) {
50         int count = 0;
51         diff = 0x3f3f3f3f;
52         gets(num);
53         for (int i = 0; i < strlen(num); i++)
54         {
55             if (num[i] >= ‘0‘ && num[i] <= ‘9‘) input[count++] = num[i] - ‘0‘;
56         }
57
58         rc = count >> 1;    //left count
59         lc = count - rc;   //right count
60         if (count % 2 == 0) {
61             int tempCount = 0;
62             do {
63                 diff = min(diff, abs(getNumber(input, lc, 1) - getNumber(input + lc, rc, 1)));
64             } while (next_permutation(input, input + count));
65         }
66         else {
67             diff = min(diff, abs(getNumber(input, lc, 1) - getNumber(input + lc, rc, 0)));
68         }
69         cout << diff << endl;
70     }
71     return 0;
72 }

总结：

• 在不知道输入个数的情况下接收，用gets()接收一行字符串，然后手动分割出来：

        gets(num);
        for (int i = 0; i < strlen(num); i++)
        {
            if (num[i] >= ‘0‘ && num[i] <= ‘9‘) input[count++] = num[i] - ‘0‘;
        }

• 第一行输入n，后面要输入数据，n后面的换行符可以用cin.ignore()忽略。