Codeforces Gym 101473D Folding Machine DFS/XJBS

http://codeforces.com/gym/101473/attachments

Problem D

Folding Machine

File: machine.[c|cpp|java]

One of the main tools of a Turing machine, which allows its computing power to be bigger than

other simpler models, is an infinite tape, divided in cells, where information is stored.

A Folding machine is a machine inspired by a Turing machine. In a Folding machine, the tape

is finite, the data are integers and instead of having the functionality of the original Turing machine,

this machine uses folding tape operations.

To perform a folding operation, the machine chooses a position between adjacent cells and folds

the tape, adding the values of overlapping cells, as can seen in the figure below.
5

dobra

6 23 8 19 7 10 5 6 23 8 5 16 30 27

6+10

23+7

19+8
Notice that the machine can also fold the tape before the tape center, as shown in the next figure.

The machine can also choose to fold at the tape start or at the tape end, actually inverting the tape.
1

dobra

2 3 4 5 6 7 1 2 7 6 5 5 5

1+4

2+3

7 6 5 4 3
Science of Bends Company is developing commercial versions of their Folding machine and its

production have recently raised. The last lot produced, unfortunately, have some issues and some

machines aren’t working properly. Some additional testing is therefore needed, to avoid selling defective

machines, which would denigrate the company’s image.

To test these machines, a set of tests and tapes are given. For each tape, the machine returns some

computation result. Therefore, the engineers responsible for testing take note of the results and can

verify if they are correct. But these engineers forgot to take note of which computation was made in

each test case.

To avoid re-testing all machines again, the engineers agreed that any combination of foldings is

sound and accepted if, from a given input, it generates the expected output. You were hired to develop

a program which, given the input and output tapes, determines whether there is a folding sequence

that, starting from the input tape, generates the output tape.

把一个数字序列反复折叠，折叠后重叠的位置的数字变为所有重复的数字和。问能否把所给序列折成所给的样子。

暴力dfs。

模拟过程挺复杂，调一调还是可以写出来的。

#include <cstdio>
#include <iostream>
#include <string.h>
#include <string>
#include <map>
#include <queue>
#include <vector>
#include <set>
#include <algorithm>
#include <math.h>
#include <cmath>
#include <stack>
#define mem0(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,0x3f,sizeof(a))
using namespace std;
typedef long long ll;
typedef long double ld;
const int maxn=25,inf=0x3f3f3f3f;
const ll llinf=0x3f3f3f3f3f3f3f3f;
const ld pi=acos(-1.0L);
int a[maxn],b[maxn];
int p[maxn][maxn];
int m,n;

bool check(int step) {
for (int i=0;i<m;i++) {
if (p[step][i]!=b[i]) return false;
}
return true;
}

bool inverse(int step) {
for (int i=0;i<m;i++) {
if (p[step][m-i-1]!=b[i]) return false;
}
return true;
}

bool dfs(int len,int step) {
if (len<m) return false;
if (len==m)
if (check(step-1)||inverse(step-1)) return true; else return false;
bool z=false;
int i;
for (i=1;i<=len/2;i++) {
int lc=i-1,rc=i;
while (lc>=0&&rc<len) {
p[step][lc]=p[step-1][lc]+p[step-1][rc];
lc--;rc++;
}
for (int j=rc;j<len;j++) p[step][j-i]=p[step-1][j];
// for (int j=0;j<len-i;j++) printf("%d ",p[step][j]);
// printf("\n");
z=z|dfs(len-i,step+1);
}
for (i=len/2+1;i<len;i++) {
int lc=i-1,rc=i;
while (lc>=0&&rc<len) {
p[step][lc]=p[step-1][lc]+p[step-1][rc];
lc--;rc++;
}
for (int j=0;j<=lc;j++) p[step][j]=p[step-1][j];
// for (int j=0;j<i;j++) printf("%d ",p[step][j]);
// printf("\n");
z=z|dfs(i,step+1);
}
return z;
}

int main() {
// freopen("F.in","r",stdin);
// freopen("F.out","w",stdout);
scanf("%d",&n);
int i;
for (i=0;i<n;i++) {
scanf("%d",&a[i]);
p[0][i]=a[i];
}
scanf("%d",&m);
for (i=0;i<m;i++) {
scanf("%d",&b[i]);
}
if (dfs(n,1)) printf("S"); else printf("N");
return 0;
}