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f[i][j][S]考虑到第i列有j列全是男生,第i列的情况为S的方案数矩乘优化即可
是我的矩乘太丑了嘛?为什么跑的这么慢?
#include<cstdio> #include<cstring> #include<cstdlib> #include<cmath> #include<iostream> #include<algorithm> #define mod 1000000007 using namespace std; struct yts1 { int x,y,z; }q[410]; struct yts { int x,y; int a[210][210]; }a,b,c,ans; long long m; int CC[10][10]; int n,P,Q,size; void mul(yts &a,yts &b,yts &c) { memset(ans.a,0,sizeof(ans.a)); for (int i=1;i<=a.x;i++) for (int k=1;k<=a.y;k++) if (a.a[i][k]) for (int j=1;j<=b.y;j++) if (b.a[k][j]) ans.a[i][j]=(ans.a[i][j]+(long long)a.a[i][k]*b.a[k][j]%mod)%mod; c.x=a.x;c.y=b.y; for (int i=1;i<=c.x;i++) for (int j=1;j<=c.y;j++) c.a[i][j]=ans.a[i][j]; } void power(yts &a,long long y) { c.x=size;c.y=size; memset(c.a,0,sizeof(c.a)); for (int i=1;i<=size;i++) c.a[i][i]=1; while (y) { if (y&1) mul(c,a,c); mul(a,a,a); y>>=1; } } int main() { scanf("%d%lld%d%d",&n,&m,&P,&Q); CC[0][0]=1; for (int i=1;i<=n;i++) { CC[i][0]=1; for (int j=1;j<=i;j++) CC[i][j]=(CC[i-1][j]+CC[i-1][j-1])%mod; } memset(b.a,0,sizeof(b.a)); for (int i=0;i<=Q;i++) for (int j=0;j<=n;j++) for (int k=0;k<=n;k++) if (j+k<=n) { int z=n-j-k; if (P==1 && (z || k)) continue; if (P==2 && z) continue; q[++size]=(yts1){i,j,k}; if (i==0 && j==n && k==0) b.a[1][size]=1; } b.x=1;b.y=size;a.x=size;a.y=size; for (int i=1;i<=size;i++) for (int j=1;j<=size;j++) { int x=q[i].x,A=q[i].y,B=q[i].z,C=n-A-B; int xx=q[j].x,aa=q[j].y,bb=q[j].z,cc=n-aa-bb; if (xx<x || xx>x+1) continue; if (xx==x+1) { if (aa || A!=bb || B!=cc || C) continue; a.a[i][j]=1; } else { if (bb>A || cc>B) continue; a.a[i][j]=CC[A][bb]*CC[B][cc]%mod; if (!aa && A==bb && B==cc && !C) a.a[i][j]--; } } power(a,m); mul(b,c,b); int ans=0; for (int i=1;i<=size;i++) ans=(ans+b.a[1][i])%mod; printf("%d\n",ans); return 0; }
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原文地址:http://blog.csdn.net/u012288458/article/details/51356246