poj1090 Chain

Posted 2020-08-11 MashiroSky

http://poj.org/problem?id=1090 (题目链接)

题意

　　给出九连环的初始状态，要求将环全部取下需要走多少步。

Solution

　　格雷码：神犇博客

　　当然递推也可以做。

代码

// poj1090
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<set>
#define MOD 1000000007
#define inf 2147483640
#define LL long long
#define free(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout);
using namespace std;
inline LL getint() {
    LL x=0,f=1;char ch=getchar();
    while (ch>\'9\' || ch<\'0\') {if (ch==\'-\') f=-1;ch=getchar();}
    while (ch>=\'0\' && ch<=\'9\') {x=x*10+ch-\'0\';ch=getchar();}
    return x*f;
}

int n,a[2000],f[1010][400],t[400];

int main() {
    scanf("%d",&n);
    int sum=0;
    for (int i=1;i<=n;i++) {
        scanf("%d",&a[i]);
        sum+=a[i];
    }
    for (int i=1;i<=n;i++) {
        sum-=a[i];
        if (sum&1) a[i]=!a[i];
    }
    f[1][1]=1;
    for (int i=2;i<=1000;i++) {
        for (int j=1;j<=399;j++) f[i][j]=f[i-1][j]*2;
        for (int j=1;j<=399;j++) {
            f[i][j]+=f[i][j-1]/10;
            f[i][j-1]%=10;
        }
    }
    for (int i=1;i<=n;i++)
        if (a[i]) {
            for (int j=1;j<=399;j++) t[j]+=f[i][j];
            for (int j=1;j<=399;j++) {
                t[j]+=t[j-1]/10;
                t[j-1]%=10;
            }
        }
    int i;
    for (i=399;i>=1;i--) if (t[i]) break;
    if (i==0) printf("0");
    else for (;i>=1;i--) printf("%d",t[i]);
    return 0;
}