AC，∑ndless

562 - Dividing coinsTime limit: 3.000 secondshttp://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=503It's commonly known that the Dutch

http://poj.org/problem?id=2184dp[s]表示当TS=s时，TF的最大值。if (s > 0){ for (j = Max; j >= Min; --j) dp[j + s] = max(dp[j + s], dp[j] + f);///dp[s]表示当TS=s时，TF的最大值 Max += s;}else{ if (f <=

http://acm.hdu.edu.cn/showproblem.php?pid=2955如何变形呢？？int _01pack(int n, int maxw){ memset(dp, 0, sizeof(dp)); dp[0] = 1.0;///初始化这个 int i, j; for (i = 0; i < n; ++i) for (j = maxw; j >=

http://acm.hdu.edu.cn/showproblem.php?pid=2546先给菜价排个序，然后：printf("%d\n", m 完整代码：/*62ms,240KB*/#include#include#includeusing namespace std;const int mx = 1005;int w[mx], dp[mx];

http://acm.hdu.edu.cn/showproblem.php?pid=2602/*31ms,240KB*/#include#include#includeusing namespace std;const int mx = 1005;int w[mx], v[mx], dp[mx];void _01pack(int n, int maxw){ me

http://poj.org/problem?id=3211既然两个人可以同时洗同色的衣服，那么对每种颜色进行分析，转化为在sumtime/2的时间内能洗至多多久的衣服，则洗这种颜色的衣服至少用时sumtime-01pack(sumtime/2)完整代码：/*32ms,968KB*/#include#include#include#includeusing

http://poj.org/problem?id=1014参考了http://poj.org/showmessage?message_id=173435完整代码： /*0ms,376KB*/#include#includeconst int mx = 7;const int mxw = 1300;int m[mx], dp[mxw], deq[mxw],

http://poj.org/problem?id=1276模板题。完整代码：/*110ms,4316KB*/#include#includeconst int mx = 1005;const int mxw = 1000005;int w[mx], m[mx], dp[mxw], deq[mxw], deqv[mxw];int solve(int m

http://acm.hdu.edu.cn/showproblem.php?pid=2159参考《背包九讲》第5讲注意循环时，每次先杀一只怪，再选择不同的忍耐消耗。完整代码：/*0ms,276KB*/#include#include#includeusing namespace std;const int maxn = 105;int exp[max

A. Cut Ribbonhttp://codeforces.com/problemset/problem/189/Atime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output

思路：从上往下分析，嗯。。。一个五叉树。所以干脆从下往上递推：dp[i] = dp[i] + dp[i - coin[k]]完整代码：

147 - DollarsTime limit: 3.000 seconds http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=83和UVa 357一样。注意所有数除以5再算。完整代码：/*0.019s*/#in

http://acm.hdu.edu.cn/showproblem.php?pid=2639参考背包九讲9.5节，注意合并单调队列的技巧(末尾加一个-1标记)完整代码：/*109ms,488KB*/#include#include#include#includeusing namespace std;int w[105], v[105], dp[1005

357 - Let Me Count The WaysTime limit: 3.000 seconds http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=114&page=show_problem&problem=293思路：dp[i] += dp[i - coin[

介绍：完全背包：有N种物品和一个容量为V的背包，每种物品都有无限件可用。放入第i种物品的价值是Wi，所占空间是Ci。求解：将哪些物品装入背包，可使这些物品体积总和不超过背包容量，且价值总和最大。题目可以转化为在Σ(i=1~N) kiCi 思路：temp = f[j - c[i]] + w[i];if (f[j] < temp) f[j] = temp;读

624 - CDTime limit: 3.000 secondshttp://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=565You have a long drive by car ahead. You have a

10664 - LuggageTime limit: 3.000 secondshttp://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=1605Peter and his friends are on holiday, s

10130 - SuperSaleTime limit: 3.000 secondshttp://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=1071There is a SuperSale in a SuperHiperM

http://acm.hdu.edu.cn/showproblem.php?pid=2126由于身上的钱相当于背包容量，价格相当于重量，由于我们要买尽量多的物品，所以每件物品的价值就为1在价值都相等的情况下，我们可以用贪心的思想求出最多能买多少件物品——排个序就行了最后就是怎么就最优解的方案数，如下：for (j = 0; j <= maxw; ++j) selection