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+10;
const int maxk=40+1;
const int inf=1e9;
const ll mod=1e9+7;
const double eps=1e-8;

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

ll C(int x,int y)
{
    if (x<0)
    {
        x=y-1-x;
        return (y&1?-1:1)*mul[x]*chu[x-y]%mod*chu[y]%mod;
    }
    if (x<y)
        return 0;
    return mul[x]*chu[x-y]%mod*chu[y]%mod;
}

ll _pow(ll a,ll b)
{
    ll y=1;
    while (b)
    {
        if (b &1)
            y=y*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    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=0;
    while (x)
    {
        sum=(sum+f[j][x])%mod;
        x-=x&-x;
    }
    return sum;
}

int main()
{
    int m,i,j,mode,x;
    ll sum=0,y;
    scanf("%d%d%d",&n,&m,&k);
    k--;
    mul[0]=1;
    for (i=1;i<=n;i++)
        mul[i]=mul[i-1]*i%mod;
    chu[n]=_pow(mul[n],mod-2);
    for (i=n-1;i>=0;i--)
        chu[i]=chu[i+1]*(i+1)%mod;
    while (m--)
    {
        scanf("%d",&mode);
        if (mode)
        {
            scanf("%d",&x);
            sum=0;
            for (j=0;j<=k;j++)
                sum=(sum+C(x,j)*cal(x,j))%mod;
            printf("%lld
",sum);
        }
        else
        {
            scanf("%d%lld",&x,&y);
            for (j=0;j<=k;j++)
                update(x,C(k-x,k-j)*y%mod,j);
        }
    }
    return 0;
}
/*
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+10;
const int inf=1e9;
const double eps=1e-8;
const ll mod=1e9+7;

ll mul[maxn<<1],chu[maxn<<1];

ll _pow(ll a,ll b)
{
    ll y=1;
    while (b)
    {
        if (b&1)
            y=y*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    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[0]=1;
    for (i=1;i<=2*n;i++)
        mul[i]=mul[i-1]*i%mod;
    chu[2*n]=_pow(mul[2*n],mod-2);
    for (i=2*n-1;i>=0;i--)
        chu[i]=chu[i+1]*(i+1)%mod;
    i=0,j=0;
    while (~scanf("%d",&a))
    {
        printf("%lld
",_pow(2,1+i+j)*C(2*n-2-i-j,n-i-1)%mod);
        i+=(a==0),j+=(a==1);
    }
    return 0;
}
/*

*/

GLongest Common Subsequence