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;
}