study from : https://blog.csdn.net/mitsuha_/article/details/81774727
k较小。分离x和k。
另外的可能：求a[k][x]，x不确定，想到的是求sum(a[k][1]+a[k][2]+...+a[k][x])，树状数组 sum(x)-sum(x-1)。这题用不上。
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cmath> 4 #include <cstring> 5 #include <string> 6 #include <algorithm> 7 #include <set> 8 #include <map> 9 #include <queue> 10 #include <iostream> 11 using namespace std; 12 13 #define ll long long 14 15 const int maxn=1e5+10; 16 const int maxk=40+1; 17 const int inf=1e9; 18 const ll mod=1e9+7; 19 const double eps=1e-8; 20 21 ll f[maxk][maxn],mul[maxn],chu[maxn]; 22 int n,k; 23 24 ll C(int x,int y) 25 { 26 if (x<0) 27 { 28 x=y-1-x; 29 return (y&1?-1:1)*mul[x]*chu[x-y]%mod*chu[y]%mod; 30 } 31 if (x<y) 32 return 0; 33 return mul[x]*chu[x-y]%mod*chu[y]%mod; 34 } 35 36 ll _pow(ll a,ll b) 37 { 38 ll y=1; 39 while (b) 40 { 41 if (b &1) 42 y=y*a%mod; 43 a=a*a%mod; 44 b>>=1; 45 } 46 return y; 47 } 48 49 void update(int x,ll y,int j) 50 { 51 while (x<=n) 52 { 53 f[j][x]=(f[j][x]+y)%mod; 54 x+=x&-x; 55 } 56 } 57 58 ll cal(int x,int j) 59 { 60 ll sum=0; 61 while (x) 62 { 63 sum=(sum+f[j][x])%mod; 64 x-=x&-x; 65 } 66 return sum; 67 } 68 69 int main() 70 { 71 int m,i,j,mode,x; 72 ll sum=0,y; 73 scanf("%d%d%d",&n,&m,&k); 74 k--; 75 mul[0]=1; 76 for (i=1;i<=n;i++) 77 mul[i]=mul[i-1]*i%mod; 78 chu[n]=_pow(mul[n],mod-2); 79 for (i=n-1;i>=0;i--) 80 chu[i]=chu[i+1]*(i+1)%mod; 81 while (m--) 82 { 83 scanf("%d",&mode); 84 if (mode) 85 { 86 scanf("%d",&x); 87 sum=0; 88 for (j=0;j<=k;j++) 89 sum=(sum+C(x,j)*cal(x,j))%mod; 90 printf("%lld ",sum); 91 } 92 else 93 { 94 scanf("%d%lld",&x,&y); 95 for (j=0;j<=k;j++) 96 update(x,C(k-x,k-j)*y%mod,j); 97 } 98 } 99 return 0; 100 } 101 /* 102 4 11 3 103 0 3 1 104 1 4 105 */
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cmath> 4 #include <cstring> 5 #include <string> 6 #include <algorithm> 7 #include <set> 8 #include <map> 9 #include <queue> 10 #include <iostream> 11 using namespace std; 12 13 #define ll long long 14 15 const int maxn=1e5+10; 16 const int inf=1e9; 17 const double eps=1e-8; 18 const ll mod=1e9+7; 19 20 ll mul[maxn<<1],chu[maxn<<1]; 21 22 ll _pow(ll a,ll b) 23 { 24 ll y=1; 25 while (b) 26 { 27 if (b&1) 28 y=y*a%mod; 29 a=a*a%mod; 30 b>>=1; 31 } 32 return y; 33 } 34 35 ll C(ll x,ll y) 36 { 37 return mul[x]*chu[x-y]%mod*chu[y]%mod; 38 } 39 40 int main() 41 { 42 int n,i,j,a; 43 scanf("%d",&n); 44 mul[0]=1; 45 for (i=1;i<=2*n;i++) 46 mul[i]=mul[i-1]*i%mod; 47 chu[2*n]=_pow(mul[2*n],mod-2); 48 for (i=2*n-1;i>=0;i--) 49 chu[i]=chu[i+1]*(i+1)%mod; 50 i=0,j=0; 51 while (~scanf("%d",&a)) 52 { 53 printf("%lld ",_pow(2,1+i+j)*C(2*n-2-i-j,n-i-1)%mod); 54 i+=(a==0),j+=(a==1); 55 } 56 return 0; 57 } 58 /* 59 60 */