HDU-6333 problem B. Harvest of Apples Mo Team

HDU-6333

Test instructions: There are n different apples, you can use up to M, ask how many kinds of ways, multiple sets of data, groups and n,m are 1e5, so the table can not play.

Idea: This problem to use the combination of the properties of the number of S (n,m) from N to the maximum number of M-method total, is obviously C (n,0), C (n,1) ... C (N,m) and.

Obviously s (n,m+1) = S (n, m) + C (n,m+1);

Another equation is less obvious, s (n+1,m) = 2 * S (n,m)-C (N,M);

I also understand under the guidance of Wang Shinji.

Now that we know a group (N,M) can be transferred to (N+1,M) in a very fast time, (N-1,m), (n,m+1), (n,m-1), this time to think of Mo team. It is not magical to convert each group of N and M into the right and left endpoints of the interval.

How do you find the combination number C (n,m)? You can pre-preprocess the prefix factorial of n, each time except a bit, you can get, of course, because it is to take the meaning of the modulus, here in addition to multiply this number of inverse.

Another detail of this question is to update the right endpoint as N, in order to prevent the right endpoint from being smaller than the left endpoint, where n is smaller than M.

#include <iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<string>#include<vector>#include<map>#include<Set>#include<queue>#include<list>#include<cstdlib>#include<iterator>#include<cmath>#include<iomanip>#include<bitset>#include<cctype>#include<iostream>using namespacestd;//#pragma COMMENT (linker, "/stack:102400000,102400000")//C + +#defineLson (L, Mid, RT << 1)#defineRson (mid + 1, R, RT << 1 | 1)#defineDebug (x) cerr << #x << "=" << x << "\ n";#definePB Push_back#definePQ Priority_queuetypedefLong Longll;typedef unsignedLong LongUll;typedef pair<ll, LL >Pll;typedef pair<int,int>PII;//priority_queue<int> Q;//This is a big root heap Q//priority_queue<int,vector<int>,greater<int> >q;//This is a little Gan q#defineFi first#defineSe Second//#define Endl ' \ n '#defineOKC Ios::sync_with_stdio (false); Cin.tie (0)#defineFT (A,B,C) for (int a=b; A <= C;++a)//used to press the line#defineREP (I, J, K) for (int i = j; i < K; ++i)//Priority_queue<int, Vector<int>, greater<int> >que;Constll mos =0x7FFFFFFF;//2147483647Constll nMOS =0x80000000;//-2147483648Const intINF =0x3f3f3f3f;Constll INFF =0x3f3f3f3f3f3f3f3f;// -Template<typename t>inline T read (t&x) {x=0;intf=0;CharCh=GetChar ();  while(ch<'0'|| Ch>'9') f|= (ch=='-'), ch=GetChar ();  while(ch>='0'&&ch<='9') x=x*Ten+ch-'0', ch=GetChar (); returnx=f?-x:x;}//#define _DEBUG; //*//#ifdef _debugfreopen ("input","R", stdin);//freopen ("Output.txt", "w", stdout);#endif/*-----------------------Show Time----------------------*/                #defineBel (x) ((x-1)/b+1)Const intMAXN = 1e5+9; Const intB =233; Const intMOD = 1e9+7;                ll ANS[MAXN]; structNode {intn,m; intID;                } P[MAXN];                ll x, y; voidEXGCD (ll a,ll b) {if(b==0) {X=1; Y =0; return; } exgcd (B,a%b); LL TMP=X; X=Y; Y= tmp-a/b*Y; }                BOOLcmpConstNode &a,ConstNode &b) {                    if(BEL) = =Bel (B.M))returnA.N <B.N; returnBel (a.m.) <Bel (B.M);       } ll PM[MAXN],TWO,NY[MAXN]; //NY is preprocessing the inverse, not coming out will tle                voidinit () {pm[0] =1; EXGCD (1, MOD); ny[0] = (X + MOD)%MOD;  for(intI=1; i<maxn; i++) {Pm[i]= (pm[i-1] * i + MOD)%MOD;                        EXGCD (PM[I],MOD); Ny[i]= (X + MOD)%MOD; } EXGCD (2, MOD); both= (X + MOD)%MOD; } llGet(intNintx) {                        if(N-x <0)return 0; ll Res= (Pm[n] *ny[n-x])%MOD; Res= (res * ny[x])%MOD; returnRes; } ll sum=0; voidDel1 (intXintN) {Sum= (sum-Get(n,x) +mod)%MOD; }                voidADD1 (intXintN) {Sum= (sum +Get(n,x) +mod)%MOD; }                voidDel2 (intXintN) {Sum= ((sum +Get(n,x)) *two + MOD)%MOD; }                voidADD2 (intXintN) {Sum= (SUM *2-Get(n,x) +mod)%MOD; }intMain () {init (); intQ; scanf ("%d", &q);  for(intI=1; i<=q; i++) {scanf ("%d%d",&p[i].n,&p[i].m); P[i].id=i; } sort (P+1, p+1+q,cmp); intPL = p[1].M, PR = p[1].N;  for(intI=0; i<=pl; i++) {sum= (sum +Get(pr,i) + MOD)%MOD; } ans[p[1].id] =sum; //cout<< "* *" <<endl;                 for(intI=2; i<=q; i++){                     while(PR < P[I].N) ADD2 (PL,PR), pr++;//This is to update the right interval as n to prevent m>n;                     while(pr > P[I].N) pr--, Del2 (PL,PR);  while(PL < P[I].M) pl++, Add1 (PL,PR);  while(pl > p[i].m) del1 (PL,PR), pl--; Ans[p[i].id]= sum%MOD; }                 for(intI=1; i<=q; i++) {printf ("%lld\n", Ans[i]); }        return 0;}

HDU6333

HDU-6333 problem B. Harvest of Apples Mo Team