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# Fill the bags

 0 Please tell me the exact logic to solve this question. I've tried a lot but always end up with WA. Here is the link of the question - http://www.codechef.com/CDSU2015/problems/MHTBAG asked 31 Jan '15, 12:16 3★ar56 -3●2 accept rate: 0%

 0 It w1 is odd, the we need one n1 object to fill the last one space. If n1==0, then the point is wasted. so, if w1 is odd, then subtract 1 from it (the last space), if n1>0, add 1 to answer and reduce one from n1. Do the same for w2. Now you have two bags each with even number weight capacity. Now, each object in n2 carries 2 units of weight. So total weight is 2*n2. Let, n2=2*n2; and W=w1+w2; So now, n2 and n1 are total weight available and W is the total weight capacity. If W<=n2, it means we can fill all the weight in W by n2. So, apply { ans+=W; //As W weight is available n2-=W; W=0; } else if(W>n2) // this means W can accommodate all n2 weight and will have spare space. { W-=n2; ans+=n2; n2=0; } Do the exact same for n1 and print the value in ans. p.s. In solving with n2, if W becomes 0, it doesn't cause any problem with n1 part as we are adding and subtracting 0 from any value. answered 31 Jan '15, 15:05 893●2●11●35 accept rate: 10%
 0 Easy to understand Solution #include using namespace std; #define ll unsigned long long int main(){ ll t, w1,w2,n1,n2,W; cin>>t; while(t--){ cin>>w1>>w2>>n1>>n2; ll ans = 0; ll temp = min(w1/2+w2/2,n2); ans=2*temp; ans += min(w1+w2-2*temp, n1); cout<
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question asked: 31 Jan '15, 12:16

question was seen: 429 times

last updated: 06 Jun '17, 21:26