Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4425 Accepted Submission(s): 2698
题解:深搜一下。。。关键要开两个数组;因为自己给下一个一半的时候自己变了,所以再开个数组记录上一状态;
代码:
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
#include<map>
#include<algorithm>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define SL(x) scanf("%lld",&x)
#define PI(x) printf("%d",x)
#define PL(x) printf("%lld",x)
#define P_ printf(" ")
#define T_T while(T--)
typedef long long LL;
const int INF=0x3f3f3f3f;
int N;
int m[110],n[110];
int ans;
void dfs(int t){
if(ans)return;
if(*max_element(m,m+N)==*min_element(m,m+N)){
ans=1;
printf("%d %d\n",t,m[0]);
return ;
}
for(int i=1;i<N;i++){
m[i]=(n[i-1]+n[i])/2;
if(m[i]&1)m[i]++;
// if(m[i]&1)m[i]++;
}
m[0]=(n[0]+n[N-1])/2;
if(m[0]&1)m[0]++;
for(int i=0;i<N;i++)n[i]=m[i];
// if(m[0]&1)m[0]++;
//for(int i=0;i<N;i++)printf("%d ",m[i]);puts("");
//getchar();
dfs(t+1);
}
int main(){
while(SI(N),N!=0){
for(int i=0;i<N;i++)scanf("%d",&n[i]),m[i]=n[i];
ans=0;
dfs(0);
}
return 0;
}
原文:http://www.cnblogs.com/handsomecui/p/5023595.html