bzoj 3782: 上学路线

这和51nod 1486 那题是差不多的，只不过这里过了一个合数的取模，拆成素数然后用中国剩余定理合并起来就好

（说的好轻巧啊）

  1 /*#include<bits/stdc++.h>
  2 #define LL long long
  3 using namespace std;
  4 inline LL ra()
  5 {
  6     LL x=0; char ch=getchar();
  7     while (ch<'0' || ch>'9') ch=getchar();
  8     while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();}
  9     return x;
 10 }
 11 
 12 LL p[]={3LL,5LL,6793LL,10007LL};
 13 
 14 struct node{
 15     LL x,y;
 16 }a[500];
 17 bool cmp(node a, node b) {return a.x==b.x?a.y<b.y:a.x<b.x;}
 18 
 19 const int maxn=1000005;
 20 const int MAXN=10010;
 21 
 22 LL fac[maxn],inv[maxn];
 23 LL FAC[4][MAXN],INV[4][MAXN];
 24 LL n,m,mod,T;
 25 
 26 void pre()
 27 {
 28     if (mod==1000003)
 29     {
 30         fac[0]=1;
 31         for (int i=1; i<maxn; i++) fac[i]=fac[i-1]*i%mod; 
 32         inv[0]=inv[1]=1;
 33         for (int i=2; i<maxn; i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
 34         for (int i=2; i<maxn; i++) inv[i]=inv[i]*inv[i-1]%mod;
 35         return;
 36     }
 37     for (int j=0; j<4; j++)
 38     {
 39         FAC[j][0]=1;
 40         for (int i=1; i<MAXN; i++) FAC[j][i]=FAC[j][i-1]*i%p[j];
 41         INV[j][0]=INV[j][1]=1;
 42         for (int i=2; i<MAXN; i++) INV[j][i]=(p[j]-p[j]/i)*INV[j][p[j]%i]%p[j];
 43         for (int i=2; i<MAXN; i++) INV[j][i]=INV[j][i-1]*INV[j][i]%p[j];
 44     }
 45 //    cout<<FAC[0][2]*INV[0][1]*INV[0][1];
 46 }
 47 
 48 LL lucas1(LL n, LL m)
 49 {
 50     if (n<m) return 0;
 51     if (n<mod && m<mod) return fac[n]*inv[m]%mod*inv[n-m]%mod;
 52     return lucas1(n/mod,m/mod)*lucas1(n%mod,m%mod)%mod;
 53 }
 54 
 55 LL lucas2(LL n, LL m, int opt)
 56 {
 57     if (n<m) return 0;// cout<<n<<' '<<m<<' '<<FAC[opt][n]*INV[opt][m]%p[opt]*INV[opt][n-m]%p[opt]; system("pause");
 58     if (n<p[opt] && m<p[opt]) return FAC[opt][n]*INV[opt][m]%p[opt]*INV[opt][n-m]%p[opt];
 59     return lucas2(n/p[opt],m/p[opt],opt)*lucas2(n%p[opt],m%p[opt],opt)%p[opt];
 60 }
 61 LL ksm(LL x, int p, int mod)
 62 {
 63     LL sum=1;
 64     for (;p;p>>=1,x=x*x%mod)
 65         if (p&1) sum=sum*x%mod;
 66     return sum;
 67 }
 68 LL C(LL n, LL m)
 69 {
 70 //    printf("%d %d %d\n",n,m,lucas1(n,m));
 71     if (mod==1000003) return lucas1(n,m);
 72     else{
 73         LL ans=0;
 74         for (int i=0; i<4; i++)
 75             ans+=lucas2(n,m,i)*ksm(mod/p[i],p[i]-2,p[i])%mod*(mod/p[i])%mod,ans=(ans+mod)%mod;
 76         return ans;
 77     }
 78 }
 79 
 80 LL f[MAXN];
 81 
 82 int main()
 83 {
 84     n=ra(); m=ra(); T=ra(); mod=ra(); pre();
 85     for (int i=1; i<=T; i++) a[i].x=ra(),a[i].y=ra();
 86     a[++T].x=n; a[T].y=m;
 87     sort(a+1,a+T+1,cmp);
 88     for (int i=1; i<=T; i++)
 89     {
 90         f[i]=C(a[i].x+a[i].y,a[i].x);// cout<<a[i].x<<"  "<<a[i].y<<"   "<<f[i]<<endl;
 91         for (int j=1; j<i; j++)
 92             if (a[j].y<=a[i].y)
 93             f[i]=(f[i]-C(a[i].x-a[j].x+a[i].y-a[j].y,a[i].x-a[j].x)*f[j]%mod+mod)%mod;
 94     }
 95     cout<<f[T]<<endl;
 96     return 0;
 97 }*/
 98 
 99 #include<bits/stdc++.h>
100 #define LL long long
101 using namespace std;
102 struct node{
103     LL x,y;
104 }a[500];
105 bool cmp(node a, node b){
106     return a.x==b.x?a.y<b.y:a.x<b.x;
107 }
108 LL n,m,P;  
109 LL p[10];  
110 LL fac[5][1000010],inv[5][1000010];  
111 LL f[1010];  
112 LL g[10],sum[10],INV[10];  
113 int num;  
114 bool w=0;  
115 LL ksm(LL x, LL y, LL p)
116 {
117     LL sum=1;
118     for (;y;y>>=1,x=x*x%P)
119         if (y&1LL) sum=sum*x%P;
120     return sum;
121 }
122 LL c(LL x, LL y, int i)
123 {
124     if (x<p[i] && y<p[i]) return fac[i][y]*inv[i][x]%p[i]*inv[i][y-x]%p[i];
125     else return c(x%p[i],y%p[i],i)*c(x/p[i],y/p[i],i)%p[i]; 
126 }
127 LL C(LL y, LL x)
128 {
129     if (!w) return c(x,y,0);
130     else
131     {
132         for (int i=1; i<=4; i++) g[i]=c(x,y,i);
133         LL ans=0;
134         for (int i=1; i<=4; i++) ans=(ans+(g[i]*sum[i])%P*INV[i]%P)%P;
135         return ans;
136     }
137 }
138 int main()
139 {
140     scanf("%lld%lld%lld%lld",&n,&m,&num,&P);
141     for (int i=1; i<=num; i++) scanf("%lld%lld",&a[i].x,&a[i].y);
142     a[++num].x=n; a[num].y=m;
143     sort(a+1,a+num+1,cmp);
144     if (P!=1000003) p[1]=3,p[2]=5,p[3]=6793,p[4]=10007,w=1; else p[0]=1000003;
145     if (w)
146     {
147         for (int j=1; j<=4; j++)
148         {
149             fac[j][0]=1; sum[j]=P/p[j]; INV[j]=ksm(sum[j],p[j]-2,p[j]);
150             for (int i=1; i<p[j]; i++) fac[j][i]=fac[j][i-1]*i%p[j];
151             inv[j][p[j]-1]=ksm(fac[j][p[j]-1],p[j]-2,p[j]);
152             for (int i=p[j]-2; i>=0; i--) inv[j][i]=inv[j][i+1]*(i+1)%p[j];
153         }
154     }
155     else
156     {
157         fac[0][0]=1;
158         for (int i=1; i<P; i++) fac[0][i]=fac[0][i-1]*i%P;
159         inv[0][P-1]=ksm(fac[0][P-1],P-2,P);
160         for (int i=P-2; i>=0; i--) inv[0][i]=inv[0][i+1]*(i+1)%P;
161     }
162     for (int i=1; i<=num; i++)
163     {
164         f[i]=C(a[i].x+a[i].y,a[i].x);
165         for (int j=1; j<i; j++)
166             if (a[i].y>=a[j].y)
167                 f[i]=(f[i]-f[j]*C(a[i].x-a[j].x+a[i].y-a[j].y,a[i].x-a[j].x)%P+P)%P;
168     }
169     cout<<f[num];
170     return 0;
171 }

 

posted @ 2017-04-20 09:58  ws_ccd  阅读(247)  评论(0)  编辑  收藏  举报