USACO Combination Lock 解题日志

经历了一学期头文件大作业的摧残，终于又能刷USACO了！

虽然大神们对这上面的题很不屑，但是，菜鸡如我，还是觉得很不错的。基础要打好，嗯！

先放上题目：

Combination Lock


Farmer John's cows keep escaping from his farm and causing mischief. To try and prevent them from leaving, he purchases a fancy combination lock to keep his cows from opening
the pasture gate.
Knowing that his cows are quite clever, Farmer John wants to make sure they cannot easily open the lock by simply trying many different combinations. The lock has three dials,
each numbered 1..N (1 <= N <= 100), where 1 and N are adjacent since the dials are circular. There are two combinations that open the lock, one set by Farmer John, and also a "master" combination set by the lock maker.
The lock has a small tolerance for error, however, so it will open even if the numbers on the dials are each within at most 2 positions of a valid combination.
For example, if Farmer John's combination is (1,2,3) and the master combination is (4,5,6), the lock will open if its dials are set to (1,3,5) (since this is close enough to Farmer
John's combination) or to (2,4,8) (since this is close enough to the master combination). Note that (1,5,6) would not open the lock, since it is not close enough to any one single combination.
Given Farmer John's combination and the master combination, please determine the number of distinct settings for the dials that will open the lock. Order matters, so the setting
(1,2,3) is distinct from (3,2,1).

PROGRAM NAME: combo

INPUT FORMAT:

Line 1:	The integer N.
Line 2:	Three space-separated integers, specifying Farmer John's combination.
Line 3:	Three space-separated integers, specifying the master combination (possibly the same as Farmer John's combination).

SAMPLE INPUT (file combo.in):

50
1 2 3
5 6 7

INPUT DETAILS:

Each dial is numbered 1..50. Farmer John's combination is (1,2,3), and the master combination is (5,6,7).

OUTPUT FORMAT:

Line 1:

The number of distinct dial settings that will open the lock.

SAMPLE OUTPUT (file combo.out):

249

SAMPLE OUTPUT EXPLANATION

Here's a list:

1,1,1  2,2,4  3,4,2  4,4,5  5,4,8  6,5,6  7,5,9  3,50,2  50,1,4
1,1,2  2,2,5  3,4,3  4,4,6  5,4,9  6,5,7  7,6,5  3,50,3  50,1,5
1,1,3  2,3,1  3,4,4  4,4,7  5,5,5  6,5,8  7,6,6  3,50,4  50,2,1
1,1,4  2,3,2  3,4,5  4,4,8  5,5,6  6,5,9  7,6,7  3,50,5  50,2,2
1,1,5  2,3,3  3,4,6  4,4,9  5,5,7  6,6,5  7,6,8  49,1,1  50,2,3
1,2,1  2,3,4  3,4,7  4,5,5  5,5,8  6,6,6  7,6,9  49,1,2  50,2,4
1,2,2  2,3,5  3,4,8  4,5,6  5,5,9  6,6,7  7,7,5  49,1,3  50,2,5
1,2,3  2,4,1  3,4,9  4,5,7  5,6,5  6,6,8  7,7,6  49,1,4  50,3,1
1,2,4  2,4,2  3,5,5  4,5,8  5,6,6  6,6,9  7,7,7  49,1,5  50,3,2
1,2,5  2,4,3  3,5,6  4,5,9  5,6,7  6,7,5  7,7,8  49,2,1  50,3,3
1,3,1  2,4,4  3,5,7  4,6,5  5,6,8  6,7,6  7,7,9  49,2,2  50,3,4
1,3,2  2,4,5  3,5,8  4,6,6  5,6,9  6,7,7  7,8,5  49,2,3  50,3,5
1,3,3  3,1,1  3,5,9  4,6,7  5,7,5  6,7,8  7,8,6  49,2,4  50,4,1
1,3,4  3,1,2  3,6,5  4,6,8  5,7,6  6,7,9  7,8,7  49,2,5  50,4,2
1,3,5  3,1,3  3,6,6  4,6,9  5,7,7  6,8,5  7,8,8  49,3,1  50,4,3
1,4,1  3,1,4  3,6,7  4,7,5  5,7,8  6,8,6  7,8,9  49,3,2  50,4,4
1,4,2  3,1,5  3,6,8  4,7,6  5,7,9  6,8,7  1,50,1 49,3,3  50,4,5
1,4,3  3,2,1  3,6,9  4,7,7  5,8,5  6,8,8  1,50,2 49,3,4  49,50,1
1,4,4  3,2,2  3,7,5  4,7,8  5,8,6  6,8,9  1,50,3 49,3,5  49,50,2
1,4,5  3,2,3  3,7,6  4,7,9  5,8,7  7,4,5  1,50,4 49,4,1  49,50,3
2,1,1  3,2,4  3,7,7  4,8,5  5,8,8  7,4,6  1,50,5 49,4,2  49,50,4
2,1,2  3,2,5  3,7,8  4,8,6  5,8,9  7,4,7  2,50,1 49,4,3  49,50,5
2,1,3  3,3,1  3,7,9  4,8,7  6,4,5  7,4,8  2,50,2 49,4,4  50,50,1
2,1,4  3,3,2  3,8,5  4,8,8  6,4,6  7,4,9  2,50,3 49,4,5  50,50,2
2,1,5  3,3,3  3,8,6  4,8,9  6,4,7  7,5,5  2,50,4 50,1,1  50,50,3
2,2,1  3,3,4  3,8,7  5,4,5  6,4,8  7,5,6  2,50,5 50,1,2  50,50,4
2,2,2  3,3,5  3,8,8  5,4,6  6,4,9  7,5,7  3,50,1 50,1,3  50,50,5
2,2,3  3,4,1  3,8,9  5,4,7  6,5,5  7,5,8

其实这道题就是简单的排列组合。一开始我以为要用什么高深的东西，害怕会超内存啊超时什么的。后来发现，答案最多也才 5 * 5 * 5 * 2 = 250 种，所以一一枚举就好了。

值得注意的是，因为锁的圆的，转圈的。因此，如果数字取模后为0，要记得把值再赋成n。

下面贴上代码：

#include<fstream>
#include<iostream>
#include<cmath>
#define maxn 101
using namespace std;

ifstream fin("combo.in");
ofstream fout("combo.out");

int farmer[3], master[3], cow[3];
int store[100100];
int n, cnt;

int get(int num)
{
int i = 0;
while (num){
num /= 10;
++i;
}

return pow(10, i);
}

int solve()
{
int total = 0;
int i, j, k;
int xtmp, ytmp, ztmp;
xtmp = (farmer[0] - 2 + n) % n;
if (xtmp == 0) xtmp = n;
for (i = 0; i < 5; ++i){
ytmp = (farmer[1] - 2 + n) % n;
if (ytmp == 0) ytmp = n;
for (j = 0; j < 5; ++j){
ztmp = (farmer[2] - 2 + n) % n;
if (ztmp == 0) ztmp = n;
for (k = 0; k < 5; ++k){
//if (xflag[xtmp] == 0 || yflag[ytmp] == 0 || zflag[ztmp] == 0){
++total;
int num = get(ztmp);
int yytmp = num * ytmp + ztmp;
int num1 = get(yytmp);
int xxtmp = num1 * xtmp + yytmp;
//cout << xxtmp << " ";
int aa;
for (aa = 0; aa < cnt; ++aa){
if (store[aa] == xxtmp){
--total;
break;
}
}
if (aa == cnt || cnt == 0) store[cnt++] = xxtmp;
//}
ztmp = (ztmp + 1) % n;
if (ztmp == 0) ztmp = n;
}
ytmp = (ytmp + 1) % n;
if (ytmp == 0) ytmp = n;
}
xtmp = (xtmp + 1) % n;
if (xtmp == 0) xtmp = n;
}
xtmp = (master[0] - 2 + n) % n;
if (xtmp == 0) xtmp = n;
for (i = 0; i < 5; ++i){
ytmp = (master[1] - 2 + n) % n;
if (ytmp == 0) ytmp = n;
for (j = 0; j < 5; ++j){
ztmp = (master[2] - 2 + n) % n;
if (ztmp == 0) ztmp = n;
for (k = 0; k < 5; ++k){
//if (xflag[xtmp] == 0 || yflag[ytmp] == 0 || zflag[ztmp] == 0){
++total;
int num = get(ztmp);
int yytmp = num * ytmp + ztmp;
int num1 = get(yytmp);
int xxtmp = num1 * xtmp + yytmp;
int aa;
for (aa = 0; aa < cnt; ++aa){
if (store[aa] == xxtmp){
--total;
break;
}
}
if (aa == cnt || cnt == 0) store[cnt++] = xxtmp;
//}
ztmp = (ztmp + 1) % n;
if (ztmp == 0) ztmp = n;
}
ytmp = (ytmp + 1) % n;
if (ytmp == 0) ytmp = n;
}
xtmp = (xtmp + 1) % n;
if (xtmp == 0) xtmp = n;
}
return total;
}

int main()
{
fin >> n;
for (int i = 0; i < 3; ++i) fin >> farmer[i];
for (int i = 0; i < 3; ++i) fin >> master[i];

fout << solve() << endl;

fin.close();
fout.close();
return 0;
}