问题 B: Harvest of Apples
时间限制: 1 Sec 内存限制: 128 MB
提交: 78 解决: 35
[提交] [状态] [讨论版] [命题人:admin]
题目描述
There are n apples on a tree, numbered from 1 to n.
Count the number of ways to pick at most m apples.
输入
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).
输出
For each test case, print an integer representing the number of ways modulo 109+7.
样例输入
2 5 2 1000 500
样例输出
16 924129523
官方题解
重点就是划出来的公式,然后套莫队即可
代码:
#include <bits/stdc++.h>
using namespace std;
const int maxn=1e5+100;
#define INF 0x3f3f3f;
const int mod=1e9+7;
typedef long long ll;
struct node
{
ll n,m;
int p;
}s[maxn];
ll block;
ll fac[maxn],inv[maxn];
ll ans[maxn];
ll noww;
ll qpow(ll x,ll y){
ll ret = 1;
while(y){
if(y&1) ret = ret * x % mod;
x = x * x % mod;
y >>= 1;
}
return ret;
}
void init(){
fac[0] = 1;
for(int i=1; i<maxn; i++) fac[i] = fac[i-1] * i % mod;
inv[maxn-1] = qpow(fac[maxn-1],mod-2);
for(int i=maxn-2; i>=0; i--) inv[i] = inv[i+1] * (i+1) % mod;
}
ll C(ll x,ll y)
{
return fac[x] * inv[y] %mod * inv[x-y] % mod;
}
bool cmp(node x,node y)
{
if(x.n / block == y.n / block) return x.m < y.m;
return x.n / block < y.n / block;
}
int main()
{
int t;
scanf("%d",&t);
block = sqrt(maxn);
init();
for(int i=0; i<t; i++){
scanf("%lld%lld",&s[i].n,&s[i].m);
s[i].p = i;
}
sort(s,s + t,cmp);
ll l = 1,r = 1;
noww = 2;
for(int i=0; i<t; i++){
while(l < s[i].n){
l++;
noww = (2 * noww % mod - C(l-1,r) + mod) % mod;
}
while(l > s[i].n){
noww = (noww + C(l-1,r)) % mod * inv[2] % mod;
l--;
}
while(r < s[i].m){
r++;
noww = (noww + C(l,r)) % mod;
}
while(r > s[i].m){
noww = (noww + mod - C(l,r)) % mod;
r--;
}
ans[s[i].p] = noww;
}
for(int i=0; i<t; i++){
printf("%lld\n",ans[i]);
}
return 0;
}
刚刚学了莫队,赶紧把原来两个莫队给补了