ZOJ

来源：互联网发布：观山湖区经济发展数据编辑：程序博客网时间：2024/06/02 05:16

sg函数打表。一开始我是这么打的表。天真以为nlogn应该可以。但是跑了很久。。。。

void gsg() {memset(sg, 0, sizeof(sg));sg[0]=0;for (int i = 1; i < 5000024; i++) {memset(Hash, 0, sizeof(Hash));for (int j = 1; j*j <= i; j++)if(i%j==0){if(j!=1)Hash[sg[j]] =Hash[sg[i/j]]=1;else if(i!=1) Hash[sg[j]]=1;}for(int j=0;;j++)if (Hash[j] == 0) {sg[i] = j;if(j>29) cout<<"1"<<endl;break;}}}

后面观察了下sg[i]，发现就是代表i中有多少个素因子。于是变成这样打表，模拟素数打表的方法

void work(){sg[0]=sg[1]=0;for(int i=2;i<5000024;i++){if(sg[i]==0) sg[i]=1;for(int j=2*i;j<5000024;j+=i){if(sg[i]+1>sg[j]) sg[j]=sg[i]+1;}}}

最终代码

#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>  #include <iostream>  #include <cstdlib>  #include <cstring>  #include <string>#include <cstdio>  #include <climits>#include <cmath> #include <vector>#include <set>#include <queue>#include <stack>#include <map>#include <sstream>#include <time.h>#define INF 0x3f3f3f3f#define LL long long#define fora(i,a,n) for(int i=a;i<=n;i++)#define fors(i,n,a) for(int i=n;i>=a;i--)#define sci(x) scanf("%d",&x)#define scl(x) scanf("%lld",&x)#define MAXN 100024const double eps = 1e-8;using namespace std;int sg[5000024];//int Hash[29];//int p[5000024];int a[100024];int k=1;void work(){sg[0]=sg[1]=0;for(int i=2;i<5000024;i++){if(sg[i]==0) sg[i]=1;for(int j=2*i;j<5000024;j+=i){if(sg[i]+1>sg[j]) sg[j]=sg[i]+1;}}}//void gsg() {//memset(sg, 0, sizeof(sg));//sg[0]=0;//for (int i = 1; i < 5000024; i++) {//memset(Hash, 0, sizeof(Hash));//for (int j = 1; j*j <= i; j++)//if(i%j==0){//if(j!=1)Hash[sg[j]] =Hash[sg[i/j]]=1;//else if(i!=1) Hash[sg[j]]=1;//}//for(int j=0;;j++)//if (Hash[j] == 0) {//sg[i] = j;//if(j>29) cout<<"1"<<endl;//break;//}//}//}int main() {#ifdef DIDfreopen("in.txt", "r", stdin);//freopen("out.txt","w",stdout);clock_t S=clock(),F;#endifwork();/*gsg();for(int i=0;i<5000024;i++)if(sg[i]!=p[i])cout<<sg[i]<<" "<<p[i]<<endl;*/int n;while (scanf("%d",&n)==1) {printf("Test #%d: ",k++);int XOR = 0;for (int i = 0; i < n; i++)sci(a[i]), XOR ^= sg[a[i]];if (XOR){printf("Alice ");int ans=-1;fora(i,0,n-1){for(int j=1;j*j<=a[i];j++)if(a[i]%j==0){if(j!=1){   if((XOR^sg[a[i]]^sg[j])==0) {ans=i;break;} //记得加括号，因为==优先级比符号^高。~~   if((XOR^sg[a[i]]^sg[a[i]/j])==0) {ans=i;break;}}else if(a[i]!=1)if((XOR^sg[a[i]]^sg[j])==0) {ans=i;break;}}if(ans!=-1) break;}printf("%d\n",ans+1);}else printf("Bob\n");}#ifdef DIDF=clock();printf("时间：%lf\n",double(F-S));#endifreturn 0;}

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