### Svada

Given below code is for svada spoj.

Hint : -> Binary search :

Here actually what I am doing is partitioning time in two part and checking

1->how many coconut can be plucked by first group of monkey in first part (let say X the number of coconut)

2-> how many coconut can be braked by second group of monkey in second part(say Y - the number of coconut)

let say for any partition of time if X <= Y then that partition is possible .

And among such possible partition you have to check for partition which have least Y (according to question)

For checking we have applied binary search on time .

Here actually what I am doing is partitioning time in two part and checking

1->how many coconut can be plucked by first group of monkey in first part (let say X the number of coconut)

2-> how many coconut can be braked by second group of monkey in second part(say Y - the number of coconut)

let say for any partition of time if X <= Y then that partition is possible .

And among such possible partition you have to check for partition which have least Y (according to question)

For checking we have applied binary search on time .

#include <bits/stdc++.h> using namespace std; #define LL long long LL solve(LL a,LL b,LL t) { LL res = (t - a) < 0 ? 0 : (t-a)/b + 1; return res; } LL calculate(LL a[] , LL b[] , LL t , int sz) { int total = 0; for(int i = 0 ; i < sz ; i++) total += solve(a[i] , b[i] , t); return total; } LL b_search(LL a[] , LL b[] , LL c[] , LL d[] , LL t ,int sa, int sc) { LL low = 1 , high = t , mid , lhalf , rhalf , ma = -1; while(low < high) { mid = (low + high)>>1; lhalf = calculate(a,b,mid,sa); rhalf = calculate(c,d,t - mid , sc); if(lhalf <= rhalf){ low = mid + 1; ma = max(mid,ma); } else if(lhalf > rhalf) high = mid; } return ma; } int main() { LL t; scanf("%lld",&t); int sa , sb; scanf("%d",&sa); LL a[100], b[100] , c[100] , d[100]; for(int i = 0 ; i < sa ; i++) scanf("%lld%lld",&a[i] , &b[i]); scanf("%d",&sb); for(int i = 0 ;i<sb ; i++) scanf("%lld%lld",&c[i] ,&d[i]); printf("%lld\n", b_search(a,b,c,d,t,sa,sb)); return 0; }