贪心 HDOJ 4726 Kia's Calculation

 

题目传送门

/*
    这题交给队友做，做了一个多小时，全排列，RE数组越界，赛后发现读题读错了，囧！
    贪心：先确定最高位的数字，然后用贪心的方法，越高位数字越大

    注意：1. Both A and B will have same number of digits 两个数字位数相同
            2. which is no larger than 10^6 不是大小，而是长度不超过1e6
*/
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
using namespace std;

const int MAXN = 1e6 + 10;
const int INF = 0x3f3f3f3f;
char s1[MAXN], s2[MAXN];

int main(void)        //HDOJ 4726    Kia's Calculation
{
    //freopen ("K.in", "r", stdin);

    int t, cas = 0;
    int cnt1[11], cnt2[11], cnt3[11];

    scanf ("%d", &t);
    while (t--)
    {
        scanf ("%s", &s1);
        scanf ("%s", &s2);

        printf ("Case #%d: ", ++cas);

        int len = strlen (s1);
        if (strcmp (s1, "0") == 0)
        {
            printf ("%s\n", s2);    continue;
        }
        else if (strcmp (s2, "0") == 0)
        {
            printf ("%s\n", s1);    continue;
        }

        memset (cnt1, 0, sizeof (cnt1));
        memset (cnt2, 0, sizeof (cnt2));
        memset (cnt3, 0, sizeof (cnt3));

        for (int i=0; i<len; ++i)
        {
            cnt1[s1[i]-'0']++;    cnt2[s2[i]-'0']++;
        }

        int ii = 1, jj = 1, mx = -1;
        for (int i=1; i<=9; ++i)
        {
            if (cnt1[i] == 0)    continue;
            for (int j=1; j<=9; ++j)
            {
                if (cnt2[j] == 0)    continue;
                int tmp = (i + j) % 10;
                if (tmp > mx)
                {
                    mx = tmp;    ii = i;    jj = j;
                }
            }
        }
        cnt1[ii]--;    cnt2[jj]--;
        if (!mx)
        {
            puts ("0");        continue;
        }

        for (int i=9; i>=0; --i)
        {
            for (int j=0; j<=9; ++j)
            {
                for (int k=0; k<=9; ++k)
                {
                    if ((j+k)%10==i && cnt1[j] && cnt2[k])
                    {
                        int tmp = min (cnt1[j], cnt2[k]);
                        cnt1[j] -= tmp;    cnt2[k] -= tmp;
                        cnt3[i] += tmp;
                    }
                }
            }
        }

        printf ("%d", mx);
        for (int i=9; i>=0; --i)
        {
            for (int j=0; j<cnt3[i]; ++j)    printf ("%d", i);
        }
        puts ("");
    }


    return 0;
}