Problem B. Harvest of Apples
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1169 Accepted Submission(s): 443
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
莫队大法好~
先找到状态转移
用s(n,m)表示C(n,0)+C(n,1)+...+C(n,m) 那么
s(n,m)=2*s(n-1,m)-c(n-1,m);
s(n,m)=s(n,m-1)+c(n,m);
也就是说知道一个s(n,m)就能O(1)求出s(n+1,m),s(n-1,m),s(n,m+1),s(n,m-1)
然后瞎文明搞
注意逆元要先预处理出来
然后,就过了(当然是赛后,赛时还不会莫队,弱小可怜无助还特别能wa)
AC代码:
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<map>
#include<cmath>
using namespace std;
const int maxn = 1e5+10;
const long long ha = 1e9+7;
struct node{
int pos, n, m;
bool friend operator < (node na, node nb){
return (na.m<nb.m||(na.m==nb.m&&na.n<nb.n));
}
} ex;
long long ans[maxn], c[maxn], inv[maxn];
int T, block;
vector<node> md[1000];
long long quickm(long long a, long long b){
long long ans=1;
while(b){
if (b&1){
ans=ans*a%ha;
}
b>>=1;
a*=a; a%=ha;
}
return ans;
}
void init()
{
c[1]=1;
for(int i=2;i<=maxn;i++)
c[i]=c[i-1]*i%ha;
inv[100000]=quickm(c[100000],ha-2);
for (int i=100000-1;i>=1;--i){
inv[i]=inv[i+1]*((long long)(i+1))%ha;
}
return ;
}
long long C(long long n, long long m){
if (n==m)
return 1;
if (m==0)
return 1;
return (((c[n]*inv[m])%ha)*inv[n-m])%ha;
}
void read(){
scanf("%d",&T);
block=sqrt(maxn);
int j,k;
for (int i=0;i<T;++i){
scanf("%d%d",&ex.n,&ex.m);
ex.pos=i;
j=ex.n/block;
md[j].push_back(ex);
}
return ;
}
void solve(){
int nn, nm;
long long nans;
for (int i=0;i<=block;++i){
if (md[i].size()==0){
continue;
}
sort(md[i].begin(),md[i].end());
nans=1; nn=i*block; nm=0;
for (int j=0;j<md[i].size();++j){
while( nn < md[i][j].n ){
nans= (2*nans+ha-C(nn,nm))%ha;
nn++;
}
while (nn > md[i][j].n){
nans = (nans+C(nn-1,nm))*inv[2]%ha;
nn--;
}
while (nm < md[i][j].m){
nm++;
nans = ( nans + C(nn,nm) )% ha;
}
ans[md[i][j].pos]=nans;
}
}
for (int i=0;i<T;++i){
printf("%lld\n",ans[i]);
}
return ;
}
int main(){
//freopen("b.in","r",stdin);
//freopen("out.txt","w",stdout);
init();
read();
solve();
return 0;
}