HDU 4869 turn the pokers (more than 2014 schools join the first 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 descriptionduring 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 to handle the results does she get. can you help her solve th Is problem?

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4869

Train of Thought: Unfortunately, I did not think about it during the competition. I thought about it as soon as it came out.

First, consider two flipped s, one flipped S (A) and one flipped S (B). If A> B is set, consider the number of numbers on the front after the flipped S (assuming the numbers are on the back at the beginning, the following uses 0 to indicate the back is up, and 1 to indicate the front is up ). After turning, the maximum number of 1 values may be a + B, and the minimum value is a-B. Further observation shows that after two flip operations, the value of 1 can be a-B at a minimum, A + B at a interval of 2, that is, a-B, a-B + 2, a-B + 4 ...... A + B-2, A + B. Now, if we add a number C, the number of 1 will be an interval, so we only need to maintain this interval. Note that there will be such a situation, such as a + B over N or a-B less than 0. In this case, we need to discuss and calculate a new interval. This is just a matter of detail. After calculating the number range [L, R] of 1, the final answer is C (n, L) + C (n, l + 2) + C (n, L + 4) + ...... C (n, R), and finally take the modulo 1e9 + 7. (C (n, m) is the number of M combinations in n.) Because N is relatively large, therefore, you cannot directly follow the formula C [N] [m] = C [n-1] [m] + C [n-1] m-1] Because C (n, m) = n! /(M! * (N-m )!), Then, n is pre-processed! And n! For the inverse element of 1e9 + 7, you can obtain the answer based on the formula.

#include <iostream>#include <string.h>#include <algorithm>#include <stdio.h>#define ll long long#define maxn 100010#define mod 1000000009using 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;}