USACO 1.3.3 Prime Cryptarithm

题目链接:1.3.3

我用的枚举法,即每产生一组数据就判断是否是所给数字里的.

AC还沾沾自喜,但一看题解,发现自己的代码真low...

在平时练习时,应该追求高效,精炼的代码,这样比赛时才能省出大量时间去做其他题!

/*
ID:wang9621
PROG:crypt1
LANG:C++
*/
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int cnt[];
int main()
{
    freopen("crypt1.in","r",stdin);
    freopen("crypt1.out","w",stdout);
    int n;
    scanf("%d",&n);
    for(int i = ;i<=n; i++) scanf("%d",&cnt[i]);
    int count = ;
    int flag = ;
    int flag1 = ;
    int flag2 = ;
    int cnt1,cnt2;
    for(int i = ; i<=n; i++)
    {
        if(cnt[i]==&&i==) continue;
        for(int j = ; j<=n; j++)
        {
            for(int k = ; k<=n; k++)
            {
                flag = cnt[i]*+cnt[j]*+cnt[k];
                for(int p = ; p<=n; p++)
                {
                    if(cnt[p]==&&p==) continue;
                    for(int q = ; q<=n; q++)
                    {
                        cnt1 = flag1 = flag*cnt[p];
                        cnt2 = flag2 = flag*cnt[q];
                        int flag11 = ;
                        int wei1 = ;
                        int flag22 = ;
                        int wei2 = ;
                        while(flag1)
                        {
                            int sou = flag1%;
                            for(int s = ; s<=n; s++)
                            {
                                if(cnt[s] == sou)
                                {
                                    flag11++;
                                    break;
                                }
                            }
                            wei1++;
                            flag1 /= ;
                        }
                        while(flag2)
                        {
                            int sou = flag2%;
                            for(int s = ; s<=n; s++)
                            {
                                if(cnt[s] == sou)
                                {
                                    flag22++;
                                    break;
                                }
                            }
                            wei2++;
                            flag2 /= ;
                        }
                        if(flag11==&&flag22==&&wei1==&&wei2==)
                        {
                            int sum = cnt1+cnt2*;
                            int cntt = ;
                            int cnttt = ;
                            while(sum)
                            {
                                int sou = sum%;
                                for(int s = ; s<=n; s++)
                                {
                                    if(cnt[s] == sou)
                                    {
                                        cntt++;
                                        break;
                                    }
                                }
                                cnttt++;
                                sum /= ;
                            }
                            if(cntt==&&cnttt==)
                            {
                                count++;
                            }
                        }
                    }
                }
            }
        }
    }
    printf("%d\n",count);
    return ;
}

我的代码

测试结果

Executing...
   Test 1: TEST OK [0.000 secs, 4184 KB]
   Test 2: TEST OK [0.000 secs, 4184 KB]
   Test 3: TEST OK [0.000 secs, 4184 KB]
   Test 4: TEST OK [0.000 secs, 4184 KB]
   Test 5: TEST OK [0.011 secs, 4184 KB]
   Test 6: TEST OK [0.000 secs, 4184 KB]
   Test 7: TEST OK [0.011 secs, 4184 KB]

All tests OK.

/*
ID:wang9621
PROG:crypt1
LANG:C++
*/
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int cnt[];
bool hash1(int x)
{
    while(x)
    {
        if(!cnt[x%]) return false;
        x /= ;
    }
    return true;
}
int main()
{
    freopen("crypt1.in","r",stdin);
    freopen("crypt1.out","w",stdout);
    int n,x;
    int count = ;
    scanf("%d",&n);
    for(int i = ;i<=n; i++)
    {
        scanf("%d",&x);
        cnt[x] = ;
    }
    for(int i = ; i<; i++)
    {
        if(hash1(i))
        {
            for(int j = ; j<; j++)
            {
                if(hash1(j))
                {
                    if(i*j<&&i*(j/)<&&i*(j%)<&&hash1(i*(j%))&&hash1(i*(j/))&&hash1(i*j))
                    {
                        count++;
                    }
                }
            }
        }
    }
    printf("%d\n",count);
    return ;
}

高效代码

测试结果

Executing...
   Test 1: TEST OK [0.000 secs, 4180 KB]
   Test 2: TEST OK [0.000 secs, 4180 KB]
   Test 3: TEST OK [0.000 secs, 4180 KB]
   Test 4: TEST OK [0.000 secs, 4180 KB]
   Test 5: TEST OK [0.000 secs, 4180 KB]
   Test 6: TEST OK [0.000 secs, 4180 KB]
   Test 7: TEST OK [0.000 secs, 4180 KB]

All tests OK.