From http://www.tuicool.com/articles/i2y2ma

POJ/3040_tuicool.cpp

@@ -0,0 +1,81 @@
+#include <iostream>
+#include<algorithm>
+#include<math.h>
+#include<stdio.h>
+#include<string.h>
+const int INF=0x3f3f3f3f;
+using namespace std;
+
+struct node
+{
+    int val,mou;//面值和数目
+} mon[25];
+int n,c;
+int need[25];//一种方案所需各面值的数目
+bool cmp(node a,node b)
+{
+    return a.val<b.val;
+}
+int main()
+{
+    int i,ans,ti,rest,lim,mday;
+
+    while(~scanf("%d%d",&n,&c))
+    {
+        ans=0;
+        lim=-1;
+        for(i=0;i<n;i++)
+            scanf("%d%d",&mon[i].val,&mon[i].mou);
+        sort(mon,mon+n,cmp);
+        for(i=n-1;i>=0;i--)
+            if(mon[i].val>=c)//如果面值大于c直接加数目就行了
+               ans+=mon[i].mou;
+            else
+            {
+                lim=i;
+                break;
+            }
+        while(1)
+        {
+            memset(need,0,sizeof need);
+            rest=c;
+            for(i=lim;i>=0;i--)//尽可能用大面值凑够
+            {
+                if(!mon[i].mou||!rest)
+                    continue;
+                ti=rest/mon[i].val;
+                ti=min(ti,mon[i].mou);
+                need[i]=ti;
+                rest-=ti*mon[i].val;
+            }
+            if(rest)//没凑够的话。只能往回找最小的面值凑够。保证损失的钱最少
+            {
+                for(i=0;i<=lim;i++)
+                {
+                    if(mon[i].val>=rest&&(mon[i].mou-need[i]))
+                    {
+                        need[i]++;
+                        rest=0;
+                        break;
+                    }
+                }
+                if(rest)
+                    break;
+            }
+            mday=INF;
+            for(i=0;i<=lim;i++)
+                if(need[i])
+                mday=min(mday,mon[i].mou/need[i]);//得出该种方案能支付的最大周数
+            ans+=mday;
+            for(i=0;i<=lim;i++)
+                mon[i].mou-=mday*need[i];
+        }
+        printf("%d\n",ans);
+    }
+    return 0;
+}
+/*
+11 
+6 1
+4 1
+*/