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……

#include <iostream>
#include <algorithm>
#include <iostream>
#include <stdio.h>
#include <string.h>

using namespace std;

int n;
int input[28];
char num[128];

long long pow10(int n)
{
    long long ret = 1;
    for (int i = 0 ;i < n; i++)
    {
        ret *= 10;
    }
    return ret;
}

int getNumber(int * addr, int count, bool isMin) {
    int res = 0;    
    if (isMin) {
        if (addr[0] == 0) {
            swap(addr[0], addr[1]);
        }
        for (int i = count - 1; i >= 0; i--) {
            res += *(addr + (count - 1 - i)) * pow10(i);
        }   
    }
    else {
        for (int i = 0; i < count; i++) {
            res += *(addr + i) * pow10(i);
        }    
    }
    return res;     
} 

int main() {
    cin >> n;
    cin.ignore(); 
    int lc;    //left count  
    int rc;   //right count

    int f;
    int diff;  
    for (int i = 0; i < n; i++) {
        int count = 0;
        diff = 0x3f3f3f3f;      
        gets(num);
        for (int i = 0; i < strlen(num); i++)
        {
            if (num[i] >= '0' && num[i] <= '9') input[count++] = num[i] - '0';
        }
        
        rc = count >> 1;    //left count  
        lc = count - rc;   //right count
        if (count % 2 == 0) {
            int tempCount = 0;
            do {
                diff = min(diff, abs(getNumber(input, lc, 1) - getNumber(input + lc, rc, 1)));
            } while (next_permutation(input, input + count));
        }
        else {            
            diff = min(diff, abs(getNumber(input, lc, 1) - getNumber(input + lc, rc, 0)));
        }
        cout << diff << endl;
    }  
    return 0;
}

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。