Problem B. Harvest of Apples
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 647 Accepted Submission(s): 235
Problem Description
There are n apples on a tree, numbered from 1 to n.
Count the number of ways to pick at most m apples.
Input
The first line of the input contains an integer T (1≤T≤105) denoting the number of test cases.
Each test case consists of one line with two integers n,m (1≤m≤n≤105).
Output
For each test case, print an integer representing the number of ways modulo 109+7.
Sample Input
2 5 2 1000 500
Sample Output
16 924129523
思路:
S(n+1,m)=2*S(n,m)-C(n,m);S(n,m+1)=S(n,m)+C(n,m+1)。
离线,莫对算法处理。
代码:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll mod=1e9+7;
const int maxn=1e5+10;
ll fac[maxn],inv[maxn],ans[maxn];
struct node
{
int n,m,id;
};
vector<node>G[maxn];
bool cmp(const node &a,const node &b){return a.n<b.n;}
ll pow1(ll a,ll b)
{
ll r=1;
while(b)
{
if(b&1) r=r*a%mod;
a=a*a%mod;
b/=2;
}
return r;
}
void init()
{
fac[0]=1;
for(ll i=1;i<maxn;i++) fac[i]=fac[i-1]*i%mod;
inv[maxn-1]=pow1(fac[maxn-1],mod-2);
for(ll i=maxn-2;i>=0;i--) inv[i]=inv[i+1]*(i+1)%mod;
}
ll C(int n,int m)
{
return fac[n]*inv[m]%mod*inv[n-m]%mod;
}
int main()
{
init();
int T;scanf("%d",&T);
int unit=sqrt(maxn);
for(int i=1;i<=T;i++)
{
node e;e.id=i;
scanf("%d%d",&e.n,&e.m);
G[e.m/unit+1].push_back(e);
}
for(int i=1;i<=unit+2;i++)
{
int len=G[i].size();
if(!len)continue;
sort(G[i].begin(),G[i].end(),cmp);
int x=G[i][0].n,y=0;
ll cnt=1;
for(int j=0;j<G[i].size();j++)
{
node e=G[i][j];
while(x<e.n) cnt=(2*cnt-C(x++,y)+mod)%mod;
while(y<e.m) cnt=(cnt+C(x,++y))%mod;
while(y>e.m) cnt=(cnt-C(x,y--)+mod)%mod;
ans[e.id]=cnt;
}
}
for(int i=1;i<=T;i++)
printf("%lld\n",ans[i]);
return 0;
}