题意:
求T组 C(n,0) + C(n,1) +...C(n,m) . ( 1<=T,n,m<=1e5,)
题解:
代码:
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<math.h>
#define ls (o<<1)
#define rs ((o<<1)+1)
#define mid ((l+r)>>1)
using namespace std;
typedef long long int ll;
const int maxn = 100000;
const ll Mod = 1e9+7;
int n,m,T;
ll pow_mod(int x,int t)
{
if(t==0) return 1;
ll ans = pow_mod(x,t/2);
ans = ans*ans%Mod;
if(t%2) ans = ans*x%Mod;
return ans;
}
ll fac[maxn+5],inv[maxn+5],ans[maxn+5];
int pos[maxn+5],chunk;
struct Node
{
int n,k,i;
} Q[maxn+5];
ll C(int a,int b)
{
return 1ll * fac[a] * inv[b] % Mod * inv[a - b] % Mod;
}
bool cmp(Node n1,Node n2)
{
if(pos[n1.n]==pos[n2.n]) return n1.k < n2.k;
return n1.n < n2.n;
}
int main()
{
fac[0] = 1;
for(int i=1; i<=maxn; i++) fac[i] = fac[i-1]*i%Mod;
inv[maxn] = pow_mod(fac[maxn],Mod-2);
for(int i=maxn-1; i>=0; i--) inv[i] = inv[i+1]*(i+1)%Mod;
chunk = sqrt(maxn);
for(int i=1; i<=maxn; i++) pos[i] = i/chunk + 1;
int T;
scanf("%d",&T);
for(int _=1; _<=T; _++)
{
scanf("%d %d",&Q[_].n,&Q[_].k), Q[_].i=_;
}
sort(Q+1,Q+T+1,cmp);
int l = Q[1].n,r = -1;
ll val = 0;
ll inv2 = pow_mod(2,Mod-2);
///莫队基本 在知道 (l,r) 的答案后 能够 向 (l+1,r),(l-1,r),(l,r+1),(l,r-1) O1 时间的转移。
for(int i=1; i<=T; i++)
{
if(pos[Q[i].n]!=pos[Q[i-1].n]) l = Q[i].n , r = -1, val = 0;
while(l<Q[i].n) val = (0ll + val + val + Mod - C(l++, r)) % Mod;
while(l>Q[i].n) val = (val + C(--l,r))%Mod*inv2%Mod;
while(r<Q[i].k) val = (val + C(l, ++ r)) % Mod;
//while(r>Q[i].k) val = (val + Mod - C(l, r--)) % Mod;
ans[Q[i].i] = val;
}
for(int i=1; i<=T; i++) printf("%lld\n",ans[i]);
return 0;
}
/**
写了if(pos[Q[i].n]!=pos[Q[i-1].n]) l = Q[i].n , r = -1, val = 0;就可以不写 while(r>Q[i].k) val = (val + Mod - C(l, r--)) % Mod;
因为排序时 相同块 k 已经排好了(同理按k分块也是一样的)。
for(int i=1; i<=T; i++)
{
if(pos[Q[i].n]!=pos[Q[i-1].n]) l = Q[i].n , r = -1, val = 0;
while(l<Q[i].n) val = (0ll + val + val + Mod - C(l++, r)) % Mod;
while(l>Q[i].n) val = (val + C(--l,r))%Mod*inv2%Mod;
while(r<Q[i].k) val = (val + C(l, ++ r)) % Mod;
//while(r>Q[i].k) val = (val + Mod - C(l, r--)) % Mod;
ans[Q[i].i] = val;
}
**/