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Dollar Dayz Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 4220 Accepted: 1642 DescriptionFarmer John goes to Dollar days at the Cow Store and discovers a unlimited number of tools on sale. During his first visit, the tools is selling variously for $, $, and $. Farmer John has exactly $ to spend. He can buy 5 tools at $1 tool at $ and a additional 1 tool at $. Of course, there

Dollar Dayzcrawling in process ...crawling failedTime limit:MS Memory Limit:65536KB 64bit IO Format:%I 64D %i64u SubmitStatus Practice POJ 3181Appoint Description:System Crawler (2016-05-27) DescriptionFarmer John goes to Dollar days at the Cow Store and discovers a unlimited number of tools on sale. During his first visit, the tools is selling variously for $, $, and $. Farmer John has exactly $ to spend. He can buy 5 tools at $1 tool at $ and a additional 1 tool at $. Of course, there is othe

#include #include#include#include#include#include#include#include#includeusing namespacestd;#defineN 2100#defineMet (A, B) (Memset (A,b,sizeof (a)))typedefLong LongLL; LL dp[n][2];intMain () {intN, M; while(SCANF ("%d%d", n, m)! =EOF) { intI, J; Met (DP,0); dp[0][0] =1; for(i=1; i) for(j=i; j) {dp[j][0] + = dp[j-i][0]; dp[j][1] + = dp[j-i][1]; dp[j][1] + = dp[j][0]/1000000000000000; dp[j][0] %=1000000000000000; } if(dp[n][1]==0) printf ("%i64d\n", dp[n][0]); Else{printf (

DescriptionFarmer John goes to Dollar days at the Cow Store and discovers a unlimited number of tools on sale. During his first visit, the tools is selling variously for $, $, and $. Farmer John has exactly $ to spend. He can buy 5 tools at $1 tool at $ and a additional 1 tool at $. Of course, there is other combinations for a total of 5 different ways FJ can spend all he money on tools. Here they is:1 @ us$3 + 1 @ us$2 1 @ us$3 + 2 @ us$1 1 @ us$2 + 3 @ us$1 2 @ us$2 + 1 @ us$1 5 @ us$1Writ

money is converted to integers here +0.5! Prevent accuracy Errors! at } - return 0; -}And with scanf and printf, you just need to be so concise:1 while (scanf ("%lf", money), money!=0.00) {2 printf (" %6.2f%17lld\n", money,f[int(money*+0.5)]); 3 }And then the POJ 3181:http://poj.org/problem?id=3181.Although still use coins as background, but in fact, is an integer division, the beginning of my array opened small, the result of Re, and later opened a large array after forgetting t

+ 2 + 1, 3 + 1 + 1 + 12 + 2 + 2, 2 + 2 + 1 + 1, 2 + 1 + 1 + 1 + 11 + 1 + 1 + 1 + 1 + 1Assume that the number of F (n,m) integer n is divided, where M represents the maximum number of sequences in which N is splitConsider the boundary state:M = 1 or N = 1 has only one partition: F (All) = 1M = N: equals the number of divisions of M-1 plus 1: f (n,n) = f (n,n-1) + 1M > N: It is supposed that the maximum number of the N-divided sequence will not exceed N, so f (n,m) = f (n,n)M Using dynamic progra

POJ 3181 Dollar Dayz 01 full backpack problem, pojdayz 01 A complete backpack problem. This mainly describes the number of combinations. There are more people doing two-dimensional dp. here we can use one-dimensional dp. One-dimensional transformation equation: dp [j] = dp [j-I] + dp [j]; where I represents the weight and j represents the current backpack capacity. This means that dp [j-I] indicates the maximum combination of j-I backpack weights. If

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