POJ 3040-allowance (greedy) __poj

L-allowance Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64D &%i64u Submit Status Practice POJ 3040 appoint Description:system Crawler (2016-05-01)

Description as a reward for record milk production, farmer John has decided to start paying Bessie the cow a small weekly Allowance. FJ has a set of coins in N (1 <= N <=) different denominations, where each denomination of coin evenly divides th e Next-larger denomination (e.g., 1 cent coins, 5 cent coins, cent coins, and cent). Using the given set of coins, he would like to pay Bessie at least some given amount of money C (1 <= C <= 100,000,0 ) every week. Please help him ompute the maximum number of weeks he can pay Bessie.

Input * Line 1:two space-separated integers:n and C

* Lines 2..n+1:each line corresponds to a denomination of coin and contains two integers:the value V (1 <= v <= 10 0,000,000) of the denomination, and the number of coins B (1 <= b <= 1,000,000) of this denomation in farmer John ' s Possession.

Output * Line 1: A single Integer This is the number of weeks farmer John can pay Bessie at least C allowance

Sample Input

3 6
1
1
5 120

Sample Output

111

Hint INPUT DETAILS:
FJ would like to pay Bessie 6 cents per week. He has 1-cent coins,120 5-cent coins, and 1 10-cent coin.

OUTPUT DETAILS:
FJ can overpay Bessie with the one 10-cent coin to 1 week, then pay Bessie two 5-cent coins for ten weeks and then pay Bes Sie one 1-cent coin and one 5-cent coin for weeks.
Meaning

From big to small sort, as long as not excessive can put how much, and finally from small start find a put in can excess.

Proof of correctness, because the large is a small multiple, so large put in not excessive must be put in, because the small no matter how to take, and then more than C before will be together into this large denomination, then a large replacement must be better.

After the first step is done, then must now put in a coin, then choose the smallest and can be greater than C of also must be optimal.

AC Code:

* * This greedy comparison pit where I wrote the program is not satisfied to >c is all added a value, so WA. Examples: 4 7 9 1 6 1 1 0 3 2 ans:2/#include <iostream> #include <algorithm> #include <cstring> #include & lt;string> #include <cstdio> #include <cmath> #include <ctime> #include <cstdlib> #include
<queue> #include <vector> #include <set> using namespace std;
const int t=55000;

#define INF 0X3F3F3F3FL typedef long LL; struct node {int v,w;}

A[50];

BOOL CMP (const node& A,const node& b) {return a.v>b.v;}
	int main () {#ifdef ZSC freopen ("Input.txt", "R", stdin); #endif int n,m,i,j;
		while (~SCANF ("%d%d", &n,&m)) {int num=0,cnt=0;
		int k = 0;
		memset (A,0,sizeof (a));
			for (i=0;i<n;++i) {scanf ("%d%d", &AMP;A[I].V,&AMP;A[I].W);
			Excluding the direct can pay if (a[i].v>=m) k+=a[i].w,i--, n--;
		else cnt = A[I].W;
		
		Sort (a,a+n,cmp);
			while (cnt>0) {num = m;
				From large to small money for (i=0;i<n;++i) {if (!A[I].W) continue; if (num>0){int tmp = MIN (NUM/A[I].V,A[I].W);
					A[I].W-= tmp;
					CNT-= TMP;
				num-= TMP*A[I].V;
			} if (num<=0) break; ///From small to large money if (num>0) {for (i=n-1;i>=0;--i) {if (A[I].W) {while (NUM&GT;0&AMP;&AMP;A[I].W) n
						UM-=A[I].V,A[I].W--, cnt--;
					if (num<=0) break;
	}} if (num<=0) k++;//changed here on AC} printf ("%d\n", K);
return 0;
 }