牛客网暑期ACM多校训练营 第九场

HPrefix Sum

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。

 #include <cstdio>
 #include <cstdlib>
 #include <cmath>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <set>
 #include <map>
 #include <queue>
 #include <iostream>
 using namespace std;

 #define ll long long

 const int maxn=1e5+;
 const int maxk=+;
 const int inf=1e9;
 const ll mod=1e9+;
 const double eps=1e-;

 ll f[maxk][maxn],mul[maxn],chu[maxn];
 int n,k;

 ll C(int x,int y)
 {
     if (x<)
     {
         x=y--x;
         return (y&?-:)*mul[x]*chu[x-y]%mod*chu[y]%mod;
     }
     if (x<y)
         return ;
     return mul[x]*chu[x-y]%mod*chu[y]%mod;
 }

 ll _pow(ll a,ll b)
 {
     ll y=;
     while (b)
     {
         if (b &)
             y=y*a%mod;
         a=a*a%mod;
         b>>=;
     }
     return y;
 }

 void update(int x,ll y,int j)
 {
     while (x<=n)
     {
         f[j][x]=(f[j][x]+y)%mod;
         x+=x&-x;
     }
 }

 ll cal(int x,int j)
 {
     ll sum=;
     while (x)
     {
         sum=(sum+f[j][x])%mod;
         x-=x&-x;
     }
     return sum;
 }

 int main()
 {
     int m,i,j,mode,x;
     ll sum=,y;
     scanf("%d%d%d",&n,&m,&k);
     k--;
     mul[]=;
     for (i=;i<=n;i++)
         mul[i]=mul[i-]*i%mod;
     chu[n]=_pow(mul[n],mod-);
     for (i=n-;i>=;i--)
         chu[i]=chu[i+]*(i+)%mod;
     while (m--)
     {
         scanf("%d",&mode);
         if (mode)
         {
             scanf("%d",&x);
             sum=;
             for (j=;j<=k;j++)
                 sum=(sum+C(x,j)*cal(x,j))%mod;
             printf("%lld\n",sum);
         }
         else
         {
             scanf("%d%lld",&x,&y);
             for (j=;j<=k;j++)
                 update(x,C(k-x,k-j)*y%mod,j);
         }
     }
     return ;
 }
 /*
 4 11 3
 0 3 1
 1 4
 */

CGambling

 #include <cstdio>
 #include <cstdlib>
 #include <cmath>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <set>
 #include <map>
 #include <queue>
 #include <iostream>
 using namespace std;

 #define ll long long

 const int maxn=1e5+;
 const int inf=1e9;
 const double eps=1e-;
 const ll mod=1e9+;

 ll mul[maxn<<],chu[maxn<<];

 ll _pow(ll a,ll b)
 {
     ll y=;
     while (b)
     {
         if (b&)
             y=y*a%mod;
         a=a*a%mod;
         b>>=;
     }
     return y;
 }

 ll C(ll x,ll y)
 {
     return mul[x]*chu[x-y]%mod*chu[y]%mod;
 }

 int main()
 {
     int n,i,j,a;
     scanf("%d",&n);
     mul[]=;
     for (i=;i<=*n;i++)
         mul[i]=mul[i-]*i%mod;
     chu[*n]=_pow(mul[*n],mod-);
     for (i=*n-;i>=;i--)
         chu[i]=chu[i+]*(i+)%mod;
     i=,j=;
     while (~scanf("%d",&a))
     {
         printf("%lld\n",_pow(,+i+j)*C(*n--i-j,n-i-)%mod);
         i+=(a==),j+=(a==);
     }
     return ;
 }
 /*

 */

GLongest Common Subsequence