其他

Description ``Accordian'' Patience You are to simulate the playing of games of ``Accordian'' patience, the rules for which are as follows:Deal cards one by one in a row from left to right, not overlapping. Whenever the card matches its immediate

描述 Description  某校的慣例是在每學期的期末考試之後發放獎學金。發放的獎學金共有五種，擷取的條件各自不同： 　　1)  院士獎學金，每人8000元，期末平均成績高於80分（>80），並且在本學期內發表1篇或1篇以上論文的學生均可獲得； 　　2)  五四獎學金，每人4000元，期末平均成績高於85分（>85），並且班級評議成績高於80分（>80）的學生均可獲得； 　　3)  成績優秀獎，每人2000元，期末平均成績高於90分（>90）的學生均可獲得； 　　4)

Description Matrix Chain Multiplication Suppose you have to evaluate an expression like A*B*C*D*E where A,B,C,D and E are matrices. Since matrix multiplication is associative, the order in which multiplications are performed is arbitrary. However,

Description2007/2008 ACM International Collegiate Programming Contest University of Ulm Local ContestProblem E: ExpressionsArithmetic expressions are usually written with the operators in between the two operands (which is called infix notation).

Problem DescriptionHexadecimal is very important and useful for computer programmers. You are requested to provide a hexadecimal view for given data. The hexadecimal view is made up of one or more rows. Every row except the last one represents 16

Queues and Priority Queues are data structures which are known to most computer scientists. The Team Queue, however, is not so well known, though it occurs often in everyday life. At lunch time the queue in front of the Mensa is a team queue, for

http://acm.hdu.edu.cn/showproblem.php?pid=3033借鑒了別人的思路：設計狀態dp[i][j]代表前i組容量為j的最大價值。由於一組裡面有多個物品，所以狀態轉移可以是前一組少取一個，即dp[i-1][p-g[i][j].v]+g[i][j].w，也可以是當前組之前去過的少取一種，即dp[i][p-g[i][j].v]+g[i][j].w。　　　

http://acm.hdu.edu.cn/showproblem.php?pid=1712背包九講的思路#include <iostream>#include<cstdio>#include<cstring>using namespace std;int dp[105];struct node{ int w,v;};node g[105][105];int main(){ int n,m;

http://acm.hdu.edu.cn/showproblem.php?pid=2955本來轉化成01背包 TLE~~況且浮點數不一定是兩位小數 轉換思想將總的錢數作為背包容量，逃跑成功率為價值，dp[i][j]=max(dp[i-1][j],dp[i-1][v-c[i]]+w[i]);   dp[i]表示最大逃跑成功率， 初始化的時候d[0]=1;其餘d[i]=0;即表示搶m錢時的最大成功率#include <iostream>#include

for (i=1; i<=n; i++) for (j=m; j>=q[i]; j--)f[j]=max(f[j],f[j-p[i]]+v[i]);要保證動歸方程無後效性j-p[i]一定要比j先算算i時,最小能算到q[i]-p[i]因此以q[i]-p[i]從小到大排序#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>using

Description  Polyomino Composer A polyomino is a plane geometric figure formed by joining one or more equal squares edge to edge.- WikipediaGiven a large polyomino and a small polyomino, your task is to determine whether you can compose the large

DescriptionTEX is a typesetting language developed by Donald Knuth. It takes source text together with a few typesetting instructions and produces, one hopes, a beautiful document. Beautiful documents use double-left-quote and double-right-quote to

DescriptionWatashi is the couch of ZJU-ICPC Team and he is very kind hearted. In ZJU-ICPC summer training camp, students are divided into several groups and each day one of the groups will design some problems to hold a contest. Today students of

#include<stdio.h>#include<stdlib.h>typedef struct node{    int data;    struct node *next;}Linklist;Linklist *create(){    int i;    Linklist *head=(Linklist*)malloc(sizeof(Linklist));    Linklist *p=head,*New;    head->next=NULL;   

DescriptionJudge Ito is having a problem with people subpoenaed for jury duty giving rather lame excuses in order to avoid serving. In order to reduce the amount of time required listening to goofy excuses, Judge Ito has asked that you write a

http://acm.hdu.edu.cn/showproblem.php?pid=1203先演算法失敗的機率，找出機率最小的 用1減去就是最大的機率，本來想著用dp[ i ][ j ]表示前 i 個用 j 元的最小機率 轉移方程 dp[i][j]=min(dp[ i-1 ][ j ],,,dp[ i-1 ][ j-cost[i] ]*w[i]  )感覺空間太大~~用滾動數組~~#include <iostream>#include <cstdio>#include

#include<iostream>#include<cstdio>#include<cstring>#define inf 99999999using namespace std;int dp[100005];int c[100005],w[100005];int main(){int t;scanf("%d",&t);for(;t--;){int e,f;scanf("%d%d",&e,&f);int

http://acm.hdu.edu.cn/showproblem.php?pid=1158狀態轉移方程：dp[i][j]=min(dp[i-1][k]+extra(k,j)+pay*k);  a[i-1]<=k<=max(a);這其中還有點細節要注意 當i==1時  dp[i][j]=dp[i-1][0]+extra(0,j)+p*k;extra (i,j)

http://acm.hdu.edu.cn/showproblem.php?pid=3496  二維費用背包實現恰好m控制初始化即可#include <iostream>#include <cstdio>#include <cstring>#define inf 99999999;using namespace std;int dp[1100][110];int c[110];int v[110];int main(){ int t; scanf(

http://acm.hdu.edu.cn/showproblem.php?pid=3732先用01背包 逾時，，v,c的範圍只有11個可以轉化成多重背包。memset(dp,0,sizeof(0))~~~汗sizeof(0)wa了10次。。。#include <iostream>#include <cstring>#include <cstdio>using namespace std;int dp[200005];int num[15][15];int