【HDOJ6333】Harvest of Apples（莫队）

题意：

给定T组询问，每组有两个数字n和m，求sigma i=0..m c(n,i) 答案对1e9+7取模

T<=1e5

1<=n,m<=1e5

思路：

注意要先变n再变m，否则会因n太小有些组合数会丢失

关键点在于n的转移,m的转移谁都会

预处理数组越界要小心

#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<vector>
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef vector<int> VI;
#define fi first
#define se second 
#define MP make_pair
#define MOD 1000000007
#define N   110000

struct data{int x,y,id;}a[N];
ll ans[N+10],fac[N+10],inv[N+10],inv2;
int pos[N+10],m;
 
bool cmp(data a,data b)
{
    if(pos[a.x]==pos[b.x]) return a.y<b.y;
    return a.x<b.x;
}

int read()
{ 
   int v=0,f=1;
   char c=getchar();
   while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
   while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
   return v*f;
}

ll c(int x,int y)
{
    ll t=fac[x]*inv[y]%MOD*inv[x-y]%MOD;
    return t;
}

void solve()
{
    ll sum=0; 
    for(int i=0;i<=a[1].y;i++) sum=(sum+c(a[1].x,i))%MOD;
    int nowx=a[1].x,nowy=a[1].y; 
    ans[a[1].id]=sum;
    for(int i=2;i<=m;i++)
    {
        while(nowx>a[i].x)
        {
            sum=(sum+c(nowx-1,nowy))%MOD;
            sum=sum*inv2%MOD;
            nowx--;
          }
          while(nowx<a[i].x)
        {
            sum=(sum*2-c(nowx,nowy)+MOD)%MOD;
            nowx++;
        }
        while(nowy>a[i].y)
        {
            sum=(sum-c(nowx,nowy)+MOD)%MOD;
            nowy--;
        }
        while(nowy<a[i].y)
        {
            sum=(sum+c(nowx,nowy+1)+MOD)%MOD;
            nowy++;
        }
        ans[a[i].id]=sum;
    } 
} 
        
void init()
{
    fac[0]=1;
    for(int i=1;i<=N;i++) fac[i]=fac[i-1]*i%MOD;
    inv[0]=inv[1]=1;
    for(int i=2;i<=N;i++) inv[i]=inv[MOD%i]*(MOD-MOD/i)%MOD; 
    inv2=inv[2];
    for(int i=1;i<=N;i++) inv[i]=inv[i-1]*inv[i]%MOD; 
    int block=int(sqrt(N));
    for(int i=1;i<=N;i++) pos[i]=(i-1)/block+1; 
}
 
int main()
{
      //freopen("1.in","r",stdin);
    //freopen("1.out","w",stdout);
    scanf("%d",&m);
    for(int i=1;i<=m;i++) 
    {
        scanf("%d%d",&a[i].x,&a[i].y);
        a[i].id=i;
    }
    init();
    sort(a+1,a+m+1,cmp);
    solve();
    for(int i=1;i<=m;i++) printf("%lld
",ans[i]);
    return 0;
}