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POJ3260
poj3260
链接:点击打开链接题意:顾客去买T元的物品,有N种钱币,给出每种钱币的面额和数量,卖家每种钱币有无限个.现在顾客想让交易的钱张数最少(即找回和付出钱的张数总和最少)代码:#include #include #include #include usingnamespacestd; constintINF=0x3f3f3f3f; boolf[10150]; intdp1[10150],d
stay_accept
·
2016-02-23 22:00
POJ3260
:The Fewest Coins(混合背包)
Description Farmer John has gone to town to buy some farm supplies. Being a very efficient man, he always pays for his goods in such a way that the smallest number of coins changes hands, i.e., the
·
2015-11-09 14:57
poj
POJ3260
——背包DP(多重)——The Fewest Coins
Description Farmer John has gone to town to buy some farm supplies. Being a very efficient man, he always pays for his goods in such a way that the smallest number of coins changes hands, i.e., the
·
2015-11-07 10:44
poj
POJ3260
The Fewest Coins ——完全背包+多重背包——Pku3260
只需要对John的付款数做一次多重背包,对shopkeeper的找零钱数做一次完全背包即可。 最重要的是上界的处理。可以注意到,John的付款数最多为maxv*maxv+m,也就是24400元。同理,shopkeeper找钱最多的数目为maxv*maxv. 证明如下: 如果John的付款数大于了maxv*maxv+m,即付硬币的数目大于了maxv,根据鸽笼原理,至少有两个的和对maxv取模的
·
2015-11-02 17:31
poj
poj3260
The Fewest Coins
DescriptionFarmerJohnhasgonetotowntobuysomefarmsupplies.Beingaveryefficientman,healwayspaysforhisgoodsinsuchawaythatthesmallestnumberofcoinschangeshands,i.e.,thenumberofcoinsheusestopayplusthenumberof
Kirito_Acmer
·
2015-06-30 16:00
背包DP
【
POJ3260
】The Fewest Coins 多重背包+完全背包
A来B处买东西,价值M元,有N种钱,每种钱A有一定数量,而B有无限数量。求最少用多少张钞票可以满足交易,比如样例,A出50+25,B找5,即可满足,需要3张。A用多重背包转移状态,B用完全背包。本文的多重背包优化用的是O(n)算法,二进制转换的O(nlogn)实在懒得写了。那种可以看http://blog.csdn.net/vmurder/article/details/39472419“ 【PO
Vmurder
·
2014-09-22 10:00
on
背包
完全背包
多重背包
POJ3260
POJ3260
:The Fewest Coins(混合背包)
DescriptionFarmerJohnhasgonetotowntobuysomefarmsupplies.Beingaveryefficientman,healwayspaysforhisgoodsinsuchawaythatthesmallestnumberofcoinschangeshands,i.e.,thenumberofcoinsheusestopayplusthenumberof
libin56842
·
2013-07-26 22:00
poj
背包
poj 1252 zoj 1558 Euro Efficiency 完全背包
雷同
poj3260
//看不懂题目,百度,发现题意如此,完全背包,计算方案数//大意:六种货币,面值都小于100。用这六种货币(可加可减)//用最少的货币数组成1到100。
alfredtofu
·
2011-02-19 10:00
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