https://www.nowcoder.com/acm/contest/96/F
F. 箱庭的股市
方程可以转换为A[i][j]=A[i-1][j-1]+A[i-1][j],打表后发现，每一行的后一项减去前一项，得到的
是一个杨辉三角，因此A[i][j] = sum(C(i,k)) (0<=k<=j)
其实按照数据范围，是会超时的，但是嘛……
最主要的是求sum(C(i,k))的方法(程序注释)
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cstring> 4 #include <cmath> 5 #include <list> 6 #include <stack> 7 #include <vector> 8 #include <set> 9 #include <map> 10 #include <queue> 11 #include <algorithm> 12 #include <iostream> 13 using namespace std; 14 const long mod=1e9 + 7; 15 const long maxn=1e6; 16 17 long long f[maxn+5],a[maxn+5],b[maxn+5]; 18 19 int main() 20 { 21 long m,x,y,i,ci; 22 long long be,r; 23 24 f[0]=1; 25 a[0]=1; 26 for (i=1;i<=maxn;i++) 27 { 28 f[i]=(f[i-1]<<1)%mod; 29 a[i]=a[i-1]*i%mod; 30 } 31 32 b[maxn]=1; 33 ci=mod-1-1; 34 r=a[maxn]; 35 while (ci) 36 { 37 if ((ci & 1)==1) 38 b[maxn]=b[maxn]*r%mod; 39 ci=ci>>1; 40 r=r*r%mod; 41 } 42 /* 43 a[i+1]*b[i+1] %mod = 1 44 (a[i]*(i+1))*b[i+1] %mod = 1 45 46 a[i]*b[i] %mod = 1 47 a[i]*((i+1)*b[i+1]) %mod = 1 48 */ 49 50 for (i=maxn-1;i>=1;i--) 51 b[i]=b[i+1]*(i+1)%mod; 52 b[0]=1; 53 54 while (scanf("%ld%ld%ld%lld",&m,&x,&y,&be)!=EOF) 55 { 56 x--; 57 if (x<=y) 58 { 59 printf("%lld ",f[x]*be%mod); 60 continue; 61 } 62 r=0; 63 for (i=0;i<=y;i++) 64 r=(r+ a[x]*b[x-i]%mod*b[i]%mod )%mod; 65 printf("%lld ",r*be%mod); 66 67 //x<=1e6 y<=1e4 其实如果y*2<x ,求后面的值 68 } 69 return 0; 70 }
超时代码：
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cstring> 4 #include <cmath> 5 #include <list> 6 #include <stack> 7 #include <vector> 8 #include <set> 9 #include <map> 10 #include <queue> 11 #include <algorithm> 12 #include <iostream> 13 using namespace std; 14 const long mod=1e9 + 7; 15 const long maxn=1e4+5; 16 17 long long f[maxn],c[maxn]; 18 19 int main() 20 { 21 long m,x,y,i,ci; 22 long long be,a,r; 23 24 f[0]=1; 25 for (i=1;i<=10001;i++) 26 f[i]=(f[i-1]<<1)%mod; 27 28 for (i=1;i<=10001;i++) 29 c[i]=1; 30 for (i=0;i<=10001;i++) 31 { 32 ci=mod-1-1; 33 a=i; 34 while (ci) 35 { 36 if ((ci & 1)==1) 37 c[i]=c[i]*a%mod; 38 ci=ci>>1; 39 a=a*a%mod; 40 } 41 } 42 43 while (scanf("%ld%ld%ld%lld",&m,&x,&y,&be)!=EOF) 44 { 45 x--; 46 if (x<=y) 47 { 48 printf("%lld ",f[x]*be%mod); 49 continue; 50 } 51 r=0; 52 a=1; 53 for (i=0;i<=y;i++) 54 { 55 r=(r+a)%mod; 56 a=a*(x-i)%mod *c[i+1]%mod; 57 } 58 printf("%lld ",r*be%mod); 59 } 60 return 0; 61 }