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Codeforces Round #539 (Div. 1)

时间:2020-02-07 21:40:28      阅读:60      评论:0      收藏:0      [点我收藏+]

A - Sasha and a Bit of Relax

code:

#include <cstdio>
#include <map> 
#include <cstring>
#include <algorithm>
#define N 300006   
#define ll long long  
#define setIO(s) freopen(s".in","r",stdin)
using namespace std;
int a[N],n;      
map<int,int>mp[2];    
int main()
{ 
    // setIO("input");   
    int i,j;  
    scanf("%d",&n);  
    ll ans=0ll;  
    mp[0][0]++;   
    for(i=1;i<=n;++i) 
    {
        scanf("%d",&a[i]),a[i]^=a[i-1];    
        ans+=(ll)mp[i&1][a[i]];        
        mp[i&1][a[i]]++;   
    }
    printf("%lld\n",ans);   
    return 0;
}

  

B - Sasha and One More Name

code:

#include <cstdio>
#include <map> 
#include <cstring>
#include <algorithm>
#define N 300006   
#define ll long long  
#define setIO(s) freopen(s".in","r",stdin)
using namespace std;
char S[N],A[N];   
int n;     
int check() 
{
    int i,j,flag=0,mid=n/2;            
    for(i=2;i<=mid;++i) if(S[i]!=S[i-1]) { flag=1;  break; }    
    return flag;    
}
int check_1() 
{
    int i,j;       
    for(i=2;i<n;++i) 
    { 
        int tmp=0; 
        for(j=i;j<=n;++j) A[++tmp]=S[j];   
        for(j=1;j<i;++j)  A[++tmp]=S[j]; 
        int flag=0;    
        int mid=n/2;    
        for(j=1;j<=mid;++j) if(A[j]!=A[n-j+1]) flag=1;     
        if(!flag) 
        {            
            int flag2=0; 
            for(j=1;j<=n;++j) if(A[j]!=S[j]) flag2=1;          
            if(flag2) return 1;  
        }       
    }
    return 0;   
}
int main() 
{ 
    // setIO("input");  
    int i,j;     
    scanf("%s",S+1),n=strlen(S+1);      
    if(!check()) printf("Impossible\n");  
    else if(check_1()) printf("1\n");     
    else printf("2\n");   
    return 0;
}

  

D - Sasha and Interesting Fact from Graph Theory

code:

#include <cstdio> 
#include <algorithm> 
#include <cstring>     
#define N 2000001     
#define ll long long 
#define mod 1000000007  
#define setIO(s) freopen(s".in","r",stdin) 
using namespace std;    
int fac[N],inv[N];     
int qpow(int x,int y) 
{
    if(y<0) return qpow(x,y+mod-1);   
    int tmp=1;    
    for(;y>0;y>>=1,x=(ll)x*x%mod) 
        if(y&1) tmp=(ll)tmp*x%mod;  
    return tmp;   
}  
int C(int x,int y) 
{ 
    if(x<0||y<0||x<y) return 0;   
    return (ll)fac[x]*inv[y]%mod*inv[x-y]%mod; 
}
int main() 
{ 
    // setIO("input");   
    int i,j,n,m,a,b,ans=0;      
    fac[0]=inv[0]=1;  
    for(i=1;i<N;++i) fac[i]=(ll)fac[i-1]*i%mod,inv[i]=qpow(fac[i],mod-2);   
    scanf("%d%d%d%d",&n,&m,&a,&b);   
    for(i=0;i<=n-2;++i) 
    {                 
        (ans+=(ll)C(m-1,i)*C(n-2,i)%mod*(i+2)%mod*qpow(n,n-i-3)%mod*qpow(m,n-2-i)%mod*fac[i]%mod)%=mod;   
    }
    printf("%d\n",ans);   
    return 0;
}

  

Codeforces Round #539 (Div. 1)

原文:https://www.cnblogs.com/guangheli/p/12274648.html

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