hdu 4869 Turn the pokers (2014多校联合第一场 I)
Turn the pokers
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1265 Accepted Submission(s): 465
Problem Description
During summer vacation,Alice stay at home for a long time, with nothing to do. She went out and bought m pokers, tending to play poker. But she hated the traditional gameplay. She wants to change. She puts these pokers face down, she decided to flip poker n
times, and each time she can flip Xi pokers. She wanted to know how many the results does she get. Can you help her solve this problem?
http://acm.hdu.edu.cn/showproblem.php?pid=4869
思路:可惜了,比赛中没有想出来,一比赛出来就想到了。
首先考虑两次翻转。翻转a个和b个,最好还是设a>b。如今考虑翻转后正面朝上的个数(如果一開始都是背面朝上,以下用0表示背面朝上,1表示正面朝上)。翻转后。1的个数最大可能为a+b,最小为a-b。进一步观察可发现经过两次翻转。1的个数可取的值为最小值为a-b,最大值为a+b,且间隔为2的一个区间,也就是a-b,a-b+2,a-b+4......a+b-2,a+b。那么如今如果再加一个数C,同理可得到1的个数会是一个区间,那么我们仅仅要维护这个区间就可以。
注意会有这种情况,比方a+b超过n或a-b小于0,这个时候就要讨论算出新的区间。这个细致分情况讨论就可以。
最后算出1的个数的区间[L,R]后。最后的答案就为C(n,L)+C(n,L+2)+C(n,L+4)+......C(n,R),最后要取模1e9+7。(C(n,m)为n个数中取m个的组合数)由于这里的n比較大,所以不能直接按公式C[n][m]=C[n-1][m]+C[n-1][m-1]来做。由于C(n,m)=n!/(m!*(n-m)!),那么预处理出n!和n!关于1e9+7的逆元,再依据公式就可以得到答案。
#include <iostream> #include <string.h> #include <algorithm> #include <stdio.h> #define ll long long #define maxn 100010 #define mod 1000000009 using namespace std; ll pow(ll x,ll y) { if(y==0) return 1; ll tmp=pow(x,y/2); tmp=(tmp*tmp)%mod; if(y&1) tmp=tmp*x%mod; return tmp; } ll c[maxn],b[maxn]; void init() { c[0]=1; for(int i=1;i<=100000;i++) c[i]=(c[i-1]*i)%mod; for(int i=0;i<=100000;i++) b[i]=pow(c[i],mod-2); } ll getC(int n,int m) { if(m==0||m==n) return 1; ll tmp=c[n]*b[m]%mod*b[n-m]%mod; return tmp; } int main() { freopen("dd.txt","r",stdin); init(); int n,m; while(scanf("%d%d",&n,&m)!=EOF) { int x; int l=0,r=0; for(int i=1;i<=n;i++) { scanf("%d",&x); int rr=r; if(r+x<=m) r+=x; else { if(l+x<=m) { if((m-l-x)%2) r=m-1; else r=m; } else { r=m-(l+x-m); } } if(l-x>=0) l-=x; else { if(rr>=x) { if((rr-x)%2) l=1; else l=0; } else l=x-rr; } } ll ans=0; for(int i=l;i<=r;i+=2) { ans=(ans+getC(m,i))%mod; } printf("%I64d\n",ans); } return 0; }