2017山东省赛C题(SDUT3895逆元法求组合数)

这个题需要自己模拟,然后就能看出是杨辉三角.

当然不能直接数组打表,也不能Lucas,逆元求组合数

具体解法解释请看链接组合数取模(sdut3895+HDU3037)逆元法或Lucas

如果T%2==|w-xi|%2

result+=ci*C(T/2-dis/2,T)

#include<cstdio>#include<iostream>using namespace std;#define ll long longconst int mod=1000000007;const int maxn=100010;ll jiecheng[maxn];ll quzheng(ll a,ll b){    return a>b?a-b:b-a;}ll kuaisumi(ll a,ll b){    ll ans=1;    while(b)    {        if(b&1)        {            b--;            ans=(ans*a)%mod;        }        b=b>>1;        a=(a*a)%mod;    }    ans=ans%mod;    return ans;}ll C(ll m,ll n){    if(m>n)        return 0;    ll M=kuaisumi(jiecheng[m],mod-2);    ll NM=kuaisumi(jiecheng[n-m],mod-2);    return  (((jiecheng[n]*M)%mod*NM)%mod)%mod;}int main(){    jiecheng[0]=1;jiecheng[1]=1;    for(int i=2;i<maxn;i++)    {        jiecheng[i]=jiecheng[i-1]*i%mod;    }    ll n,T,w;    while(cin>>n>>T>>w)    {        ll flag=T%2;        ll res=0;        while(n--)        {            ll x,c;            cin>>x>>c;            ll dis=quzheng(w,x);            if(dis%2==flag&&dis<=T)            {                res+=(c*C(quzheng(T/2,dis/2),T)%mod);                res=res%mod;            }        }        res=res%mod;        cout<<res<<endl;    }    return 0;}