From d563f462682dc3d3029d1582fd2715e4b29c0a8d Mon Sep 17 00:00:00 2001
From: Kirito <1362050620@qq.com>
Date: Wed, 20 Apr 2016 21:15:34 +0800
Subject: [PATCH] From www.aiuxian.com

http://www.aiuxian.com/article/p-2049140.html
---
 POJ/3040_aiuxian.cpp | 72 ++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 72 insertions(+)
 create mode 100644 POJ/3040_aiuxian.cpp

diff --git a/POJ/3040_aiuxian.cpp b/POJ/3040_aiuxian.cpp
new file mode 100644
index 0000000..7277aea
--- /dev/null
+++ b/POJ/3040_aiuxian.cpp
@@ -0,0 +1,72 @@
+#include<stdio.h>
+#include<string.h>
+#include<iostream>
+#include<algorithm>
+using namespace std;
+const int maxn=30;
+struct node
+{
+    int x,y;//存储钱的大小和数量
+    bool operator<(const node c)const
+    {
+        return x>c.x;
+    }
+}num[maxn];
+int need[maxn],n,c,kj;
+//need[]函数是是存储需要哪些钱币的组合
+int main()
+{
+    while(scanf("%d%d",&n,&c)!=EOF)
+    {
+        for(int i=0;i<n;i++)
+            scanf("%d%d",&num[i].x,&num[i].y);
+        sort(num,num+n);//排序是第一步
+        kj=0;
+        while(1)
+        {
+            int sum=c;
+            memset(need,0,sizeof(need));
+            for(int i=0;i<n;i++)//这个for循环是用于找出钱币的组合方法
+            {
+                if(sum>0)
+                {
+                    int t=sum/num[i].x;
+                    if(t==0)
+                        continue;
+                    t=min(t,num[i].y);
+                    need[i]=t;
+                    sum-=t*num[i].x;
+                }
+            }
+
+            if(sum>0)//如果在上面的组合中没有使sum变为0,则从小到大选择最小的面值组合，这样使面值超过的最小化
+            {
+                for(int i=n-1;i>=0;i--)
+                if(num[i].y&&num[i].x>=sum)
+                {
+                    sum=0;
+                    need[i]++;
+                    break;
+                }
+            }
+            if(sum>0)//如果组合不能达到c，怎么说明搜索结束
+                break;
+            int cc=2e9;
+            for(int i=n-1;i>=0;i--)//找出这个组合成的面值中最小的数量，同时减去它们。
+            {
+                if(need[i])
+                {
+                    cc=min(cc,num[i].y/need[i]);
+                }
+            }
+            kj+=cc;
+            for(int i=0;i<n;i++)
+                if(need[i])
+            {
+                num[i].y-=cc*need[i];
+            }
+        }
+        printf("%d\n",kj);
+    }
+    return 0;
+}