ZOJ3538Arrange the Schedule

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比赛被虐了，一个都不会，这题只要是找出规律，然后要注意的的mod，MOD要用long long 这坑了好久

/*** 时间：2014-9-29** 知识点：** 题意：F(n) = 3/4 * (3^L - (-1)^L) ;   D(n) = (3^(L+1) + (-1)^L) / 4 ;*/#include<iostream>#include<cstdio>#include<cstring>#include<map>#include<set>#include<cmath>#include<queue>#include<algorithm>using namespace std;typedef unsigned long long ll;const ll mod=4*1000000007ll;const ll MOD=1000000007ll;int n,m;struct node{    int pos;    char c;}g[20];bool cmp(node a,node b){    return a.pos<b.pos;}long long pow_3(int a){    long long res=1;    while(a)    {        //cout<<a<<endl;        a--;        res*=3;        res%=mod;        int  b=1;        long long s=3;        while(a-b>=0)        {            res*=s;            res%=mod;            s*=s;            s%=mod;            a-=b;            b<<=1;        }    }    return res;}void solve(){    long long ans=1,num;    if(m==0)    {         ans=4*pow_3(n-1)%mod;        cout<<ans%MOD<<endl;return;    }else{        sort(g,g+m,cmp);        ans=pow_3(g[0].pos-1)%mod;        ans=(ans*pow_3(n-g[m-1].pos))%mod;        for(int i=1;i<m;i++)        {            int l=g[i].pos-g[i-1].pos-1;            if(g[i].c!=g[i-1].c){                    //(3^(L+1) + (-1)^L) / 4                num = (((pow_3( l+1 ) -(l&1?1:(-1))) )%mod)/4;                ans = ans * num % mod;            }else if(g[i].c==g[i-1].c){                // 3/4 * (3^L - (-1)^L)                num = ((3*(pow_3( l ) +(l&1?1:(-1))) )%mod)/4;  ;                ans = ans * num % mod;            }        }       cout<<ans%MOD<<endl;    }}int main(){    while(scanf("%d%d",&n,&m)!=EOF){        for(int i=0; i<m; i++)        {            cin>>g[i].pos>>g[i].c;        }        solve();    }    return 0;}

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