Hdoj:6333-problem B. Harvest of Apples (combinatorial math + MO team algorithm + inverse)

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=6333

Tips for solving problems:

• This problem can be said to be very wonderful, first push combinatorial mathematical formula, one of the most important formula is CN^{m = Cm_{n-1 + Cm-1n-1 This formula know Yang Hui Triangle is understood, but a look found nothing to use. However, the formula can be used as a basis for further deduction.}}
• Set sn^{m = cn1 + CN2 + CN3 + ... CnM then continue using the basic formula above can be converted into}
• sn ^{m-1 = sn m-cn m
  sn m+1 = sn m + Cn m+1
  sn+1 m = 2Sn m-cn m
  Sn m = (sn+1 m + Cn m)/2}
• These formulas do not seem to work, but know that the relationship between s can be transformed by the complexity of O (1), so that 105 queries can be taken offline, that is, with Mo team to maintain. Magic, to break the head is not to think of AH. The division in the inside of the operation with the inverse of the element to deal with, the inverse can be the first full and pre-processing out.

////┏┛┻━━━━━┛┻┓//┃┃//┃━┃//┃┳┛┗┳┃//┃┃//┃┻┃//┃┃//┗━┓┏━━━┛//┃┃ God beast bless//┃┃ Code No bug! //┃┗━━━━━━━━━┓//┃┣┓//┃┏┛//┗━┓┓┏━━━┳┓┏━┛//┃┫┫┃┫┫//┗━┻━┛┗━┻━┛#include <bits/stdc++.h>using namespacestd;Const intMAXN = 1e5+ -;Const intMOD = 1e9+7; typedefLong Longll;structQuery {ll n, M, B, POS, ans;} Q[maxn];ll unit, ans, FAC[MAXN], INV[MAXN], T, Rev2;ll Quick_pow (ll A, ll b) {ll res=1;  while(b) {if(b&1) Res= Res * A%MOD; A= A * A%MOD; b>>=1; }    returnRes;}voidGET_FAC () {//preprocessing the factorial and inverse elementsfac[0] =1; fac[1] =1;  for(intI=2; i<maxn;i++) {Fac[i]= i * fac[i-1] %MOD; }    //Magical Operationinv[maxn-1] = Quick_pow (fac[maxn-1], mod-2);  for(inti=maxn-2; i>=0; i--) {Inv[i]= inv[i+1] * (i +1) %MOD; }}BOOLcmp1 (query A, query B) {if(A.B = =b.b)returna.m. <B.M; returnA.B <b.b;}BOOLcmp2 (query A, query B) {returnA.pos <B.pos;}voidinit () {Unit=sqrt (MAXN);    GET_FAC (); scanf ("%lld", &t);  for(intI=0; i<t;i++) {scanf ("%lld%lld", &AMP;Q[I].N, &q[i].m); Q[i]. B= q[i].n/Unit; Q[i].pos=i; } Rev2= Quick_pow (2, mod-2); Sort (q, q+T, CMP1);} ll C (ll N, ll m) {//get a combination of C (n, m)ll ans = fac[n] * inv[n-m]% MOD * Inv[m]%MOD; returnans;}voidAddl (ll L, ll R) {ans= ((2* ans% MOD) + mod-c (L, R))%MOD;}voidCutl (ll L, ll R) {ans= (ans + C (L-1, r)% MOD) * Rev2%MOD;}voidAddR (ll L, ll R) {ans= (ans + C (l, r+1)) %MOD;}voidCutr (ll L, ll R) {ans= (ans + mod-c (l, R))%MOD;}intMain () {init (); ll L=1, r =1; Ans=2;  for(intI=0; i<t;i++) {//offline MO Team processing        intL =Q[I].N; intR =q[i].m;  while(L < L) Addl (l++, R);  while(L > L) cutl (l--, R);  while(R > R) Cutr (L, r--);  while(R < R) AddR (L, r++); Q[i].ans=ans; } sort (q, q+T, CMP2);  for(intI=0; i<t;i++) {printf ("%lld\n", Q[i].ans); }    return 0;}

Hdoj:6333-problem B. Harvest of Apples (combinatorial math + MO team algorithm + inverse)