04 2012 档案
摘要：#include#include#include#define N 1000005using namespace std;int position[N];int main(){ memset(position,0,sizeof(position)); position[1]=0; int next=1; int i,j; for(i=2;i<N;i++) { if(position[i]) continue; for(j=i;j<N;j+=i) { position[j]=next; } next++; } int m; while(scanf("%d",&am
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摘要：How many prime numbersTime Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3302 Accepted Submission(s): 1074Problem DescriptionGive you a lot of positive integers, just to find out how many prime numbers there are.InputThere are a lot of cases. In eac.
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摘要：公式 p+q-gcd(p,q)就是所求值#include#include#includeusing namespace std;int gcd(int n,int m){ if(m==0) return n; gcd(m,n%m);}int main(){ int m,n; while(cin>>m>>n) { cout<<m+n-gcd(m,n)<<endl; }}
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摘要：Transformations
A square pattern of size N x N (1 #include
#include
#include
#define N 15
using namespace std; char map[N][N];
char _map[N][N];
int main()
{ freopen("transform.in","r",stdin); freopen("transform.out","w",stdout); int m; while(scanf("%d&quo
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摘要：Milking CowsThree farmers rise at 5 am each morning and head for the barn to milk three cows. The first farmer begins milking his cow at time 300 (measured in seconds after 5 am) and ends at time 1000. The second farmer begins at time 700 and ends at time 1200. The third farmer begins at time 1500 a
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摘要：Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2000 Accepted Submission(s): 1423Problem Description假设有x1个字母A， x2个字母B,..... x26个字母Z，同时假设字母A的价值为1，字母B的价值为2,..... 字母Z的价值为26。那么，对于给定的字母，可以找到多少价值
#include
using namespace std; int f[55],word[27],cun[55].
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摘要：Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3001 Accepted Submission(s): 1868Problem DescriptionA friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that visits
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摘要：Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3266 Accepted Submission(s): 1519Problem DescriptionA robot has been programmed to follow the instructions in its path. Instructions for the next direction the robot is to move are laid down in a gri
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摘要：Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 2 Accepted Submission(s) : 2Font: Times New Roman | Verdana | GeorgiaFont Size: ← →Problem DescriptionEddy's interest is very extensive, recently he is interested in prime number. Eddy discover th
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摘要：Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 4 Accepted Submission(s) : 3Font: Times New Roman | Verdana | GeorgiaFont Size: ← →Problem Description哥德巴赫猜想大家都知道一点吧.我们现在不是想证明这个结论,而是想在程序语言内部能够表示的数集中,任意取出一个偶数,来寻找两个素数,使得其和等于该偶数.做好了这件实事,就能说明这个猜想是成立的.由于
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摘要：素数判定 Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 8 Accepted Submission(s) : 2Font: Times New Roman | Verdana | GeorgiaFont Size: ← →Problem Description对于表达式n^2+n+41，当n在（x,y）范围内取整数值时（包括x,y）(-39
#include
#define N 4004
using namespace std;
bool i
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摘要：计算大素数因子#include
#include
#define N 20004
using namespace std;
int isp[N];
int main()
{ int m,n,max,ans; memset(isp,0,sizeof(isp)); isp[1]=1; for(int i=2;i>m) { max=0; for(int i=0;i>n; if(isp[n]>max) { max=isp[n]; ans=n; } } cout<<ans<<endl; }...
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摘要：/*ID:nealgav1
PROG:friday
LANG:C++
*/
#include
#include
#define N 403
int week[N][7];
void search()
{ memset(week,0,sizeof(week)); int i,j,k,year,ri,count=0; bool flag=0; for(i=1900,year=0;i<1900+N;i++,year++) { if(year) {for(int zz=0;zz<7;zz++) week[year][zz]=week[year-1][zz]; ...
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摘要：简单题，建个结构体/*ID:nealgav1
PROG:gift1
LANG:C++
*/
#include
#include
#define N 2004
class people
{ public: char name[20]; int money;
};
int m;
people name[N];
int getid(char name_[])
{ for(int i=0;i<m;i++) { if(strcmp(name[i].name,name_)==0) return i; }
}
int main()
{ freopen("gift1...
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摘要：简单题，不解释，就是要注意提交格式。/*ID: nealgav1
PROG: ride
LANG: C++
*/
#include
#include
int main()
{ freopen("ride.in","r",stdin); freopen("ride.out","w",stdout); char name[10],q[10]; int i,j,lenn,lenq; while(scanf("%s%s",name,q)!=EOF) { lenn=strlen(name); lenq=s
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摘要：简单博弈就那样，懂SG函数就成，最近做的博弈都千篇一律。。。#include
#include
#define N 11110
int sg[N],s[N],m,n;
bool h[N];
void ssgg()
{ int i,j; sg[0]=0; for(i=1;i=0) { h[sg[i-j]]=1; } } for(j=0;j<N;j++) { if(h[j]==0) { sg[i]=j; break; } } }
}
int m...
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摘要：博弈题，记住SG函数值是唯一的，所以只有在n>=m时可以有多种选法所以只需在平常的博弈中加个IF（N>=m）就行了#include
#include
#define N 1111
int sg[N],s[N],m,n;
bool h[N];
void ssgg()
{ int i,j; sg[0]=0; for(i=1;i=0) { h[sg[i-j]]=1; } } for(j=0;j=m) {for(i=m;i<=n;i++) { printf("%d",i); if(i!=n) ...
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摘要：超简单题用上^就行了，详细见NIM游戏小NIM#include
int main()
{ int m,s; while(scanf("%d",&m)&&m) { int ans=0; for(int i=0;i<m;i++) { scanf("%d",&s); ans^=s; } if(ans) printf("Rabbit Win!\n"); else printf("Grass Win!\n"); }
}
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摘要：没什么好说的，博弈简单题，懂得SG改下s数组大体都一样详细讲解类试题1850#include
#include
#include
using namespace std;
int n,m,p,sg[1003],h[1003],s[20];
void Sg()
{ int i,j; for(i=0;i=0)h[sg[i-s[j]]]=1; for(j=0;j<1003;j++) if(h[j]==0) { sg[i]=j; break; } }
}
int main()
{ s[0]=1;s[1]=2; for(int ...
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摘要：#include
#include
#include
using namespace std;
int n,m,sg[1003],h[1003],s[14];
void Sg()
{ int i,j; for(i=0;i=0)h[sg[i-s[j]]]=1; for(j=0;j<1003;j++) if(h[j]==0) { sg[i]=j; break; } }
}
int main()
{ for(int i=0;i<14;i++) s[i]=1<<i; while(scanf("%d",&n)!=EOF)
...
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摘要：简单博弈，懂得SG函数就能A#include
#include
#include
using namespace std;
int main()
{ int n,m,sg[1003],h[1003]; scanf("%d",&m); while(m--) { int i,j,k; scanf("%d%d",&n,&k); for(i=0;i=0)h[sg[i-j]]=1; } for(j=0;j<1002;j++) { if(h[j]==0) {sg[i]=j;break;} ...
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摘要：SG函数的应用加上^.#include
#include
#include
#define N 10003
using namespace std;
int sg[N];
int s[103],h[103];
int main()
{ int n,m,i,j; //freopen("1.txt","w",stdout); while(scanf("%d",&n)&&n) { for(i=0;i=0) h[sg[j-s[i]]]=1; } for(i=0;i=s[i]&&!sg[j-s[i]]){
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摘要：pnpnpnpnpnnnnnnnnnpnpnpnpnpnnnnnnnnnpnpnpnpnpnnnnnnnnnpnpnpnpnpnnnnnnnnnpnpnpnpnp步骤1:将所有终结位置标记为必败点（P点）；步骤2: 将所有一步操作能进入必败点（P点）的位置标记为必胜点（N点）步骤3:如果从某个点开始的所有一步操作都只能进入必胜点（N点） ，则将该点标记为必败点（P点） ；步骤4: 如果在步骤3未能找到新的必败（P点），则算法终止；否则，返回到步骤2。由表知当n,m均为奇数是为p-必败点#include
int main()
{ int m,n; while(scanf("%d%d&q
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摘要：经典的博弈题啊，必须弄懂必败点条件。这里是对n堆的牌数去异或，如果值为0则表示必败。题目问我们第一布有哪几种方法胜利。即就是第一步能够给对手构建多少个必败点。由于一次只能对一堆排进行操作，假设我 操作第i堆牌（a张），抽出x张。那么其余n-1堆牌的异或值是固定为b.那么 （a - x）^ b == 0 时，对手必败。到此可能有人像我一样觉得必须历遍所有a求出那个值x满足条件。其实不必要由上式可知x只有唯一取值而且 一个数 与 b 异或等于 0 即表明 这个数等于b.所以反过来我们可以求出b, 令 a - x = b ;只要b满足 b int main()
{ int sum,a,b[10...
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