Square Coins Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1398 Description People in Silverland use square coins. Not only they have square shapes but also their values are square numbers. Co…

Problem Description People in Silverland use square coins. Not only they have square shapes but also their values are square numbers. Coins with values of all square numbers up to 289 (=17^2), i.e., 1-credit coins, 4-credit coins, 9-credit coins, ...…

Square Coins Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7883    Accepted Submission(s): 5332 Problem Description People in Silverland use square coins. Not only they have square shapes but…

Square Coins Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Problem Description People in Silverland use square coins. Not only they have square shapes but also their values are square numbers. Coins with values o…

pid=1398">链接:hdu 1398 题意:有17种货币,面额分别为i*i(1<=i<=17),都为无限张. 给定一个值n(n<=300),求用上述货币能使价值总和为n的方案数 分析:这题能够用母函数的思想,对300以内的值进行预处理就可以 也可用全然背包思想求300以内的方案数 母函数: #include<stdio.h> int main() { int c1[305],c2[305],i,j,k,n; for(i=0;i<=300;i++){…

欢迎参加——BestCoder周年纪念赛(高质量题目+多重奖励) Square Coins Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9728    Accepted Submission(s): 6668 Problem Description People in Silverland use square coins. Not…

Problem Description People in Silverland use square coins. Not only they have square shapes but also their values are square numbers. Coins with values of all square numbers up to 289 (=17^2), i.e., 1-credit coins, 4-credit coins, 9-credit coins, ...…

Square Coins Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9903    Accepted Submission(s): 6789 Problem Description People in Silverland use square coins. Not only they have square shapes but…

Square Coins Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8800    Accepted Submission(s): 5991 Problem Description People in Silverland use square coins. Not only they have square shapes but…

题目大意:有面值分别为.1,4,9,.......17^2的硬币无数多个.问你组成面值为n的钱的方法数. 最简单的母函数模板题: #include <cstdio> #include <cstring> using namespace std; int c1[305],c2[305],n; int main(){ while(scanf("%d",&n),n){ for(int i=0;i<=n;i++)c1[i]=1,c2[i]=0; for(i…

题意: 有17种硬币,每种的面值为编号的平方,比如 1,4,9,16.....给出一个数字,求组成这个面值有多少种组法? 思路: 用普通母函数解,主要做的就是模拟乘法,因为硬币是无限的,所以每个构造式中每一个项的系数都是1.我们只需要第n项的系数,大于n的并不需要,所以大于n的项就不用再做计算了. #include <bits/stdc++.h> using namespace std; ; int main() { freopen("input.txt", "r…

预处理出完全平方数就和普通的生成函数解整数拆分一样了 #include<iostream> #include<cstdio> using namespace std; const int N=605; int n,m,q[N],a[N],b[N]; int main() { for(int i=1;i<=18;i++) q[i]=i*i; while(scanf("%d",&m)&&m) { for(n=1;q[n+1]<=m…

Square Coins Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6416    Accepted Submission(s): 4336 Problem Description People in Silverland use square coins. Not only they have square shapes but…

Square Coins Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 11690    Accepted Submission(s): 8007 Problem Description People in Silverland use square coins. Not only they have square shapes but…

Square Coins Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 11529    Accepted Submission(s): 7897 Problem Description People in Silverland use square coins. Not only they have square shapes but…

Square Coins Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6746    Accepted Submission(s): 4554 Problem DescriptionPeople in Silverland use square coins. Not only they have square shapes but a…

Square Coins 点我 Problem Description People in Silverland use square coins. Not only they have square shapes but also their values are square numbers. Coins with values of all square numbers up to 289 (=17^2), i.e., 1-credit coins, 4-credit coins, 9-c…

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 13791    Accepted Submission(s): 9493 Problem Description People in Silverland use square coins. Not only they have square shapes but also their v…

HDOJ(HDU).3466 Dividing coins ( DP 01背包 无后效性的理解) 题意分析 要先排序,在做01背包,否则不满足无后效性,为什么呢? 等我理解了再补上. 代码总览 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define nmax 505 #define nn 505*100 using namespace std;…

Description People in Silverland use square coins. Not only they have square shapes but also their values are square numbers. Coins with values of all square numbers up to 289 (=17^2), i.e., 1-credit coins, 4-credit coins, 9-credit coins, ..., and 28…

Square Coins Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 15764    Accepted Submission(s): 10843 Problem Description People in Silverland use square coins. Not only they have square shapes bu…

Lucky Coins 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5985 Description Bob has collected a lot of coins in different kinds. He wants to know which kind of coins is lucky. He finds out a lucky kind of coins by the following way. He tosses all th…

http://acm.hdu.edu.cn/showproblem.php?pid=5079 题意: n*n网格,每个格子可以涂黑色或白色,有的格子必须涂黑色 问最大白色正方形边长分别为0,1,2,……n 的涂色方案数 令ans[i]表示最大白色正方形边长小于i的方案数 最大边长=i 的就是ans[i+1]-ans[i] 枚举sz,表示现在要求最大白色正方形边长<i的方案数 设dp[i][st] 表示前i行,状态为st的方案数 st内压缩了n-sz+1个数,其中的第j个数表示 从右往左数第j列,…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1398 看到网上的题解都是说母函数……为什么我觉得就是一个dp就好了,dp[i][j]表示只用前i种硬币,组成价值为j的价格的方案数,转移枚举第i种硬币用多少个就好了. #include<bits/stdc++.h> using namespace std; ; ][maxn]; int main() { ;i<=;i++) { ;j<=;j++) { // i*i coin dp[i]…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=1028 就是可以用任意个1.2.3....,所以式子写出来就是这样:(1+x+x^2+...)(1+x^2+x^4+...)(1+x^3+x^6+...)...(1+x^n+x^(2*n)+...)... 因为求 x^n 系数,所以再往后的式子就没有贡献了,求到第 n 个式子即可. 一个x^2就像一条边一样,可以让第 k 项的系数转移给第 k+2 项.按这个思路写代码就行了. #include<iostr…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=1028 整数划分,每个数可以用无限次: 所以构造 f(x) = (1+x+x2+x3+...)(1+x2+x4+...)(1+x3+x6+...)...(1+xn) 乘起来后的 xn 的系数就是方案数: 用两个数组做即可,从第一个括号开始一个一个乘,f[i] 表示此时 xi 项的系数,后面每乘过来一个括号,相当于多了一种转移,所以加上. 代码如下: #include<iostream> #include…

Problem Description A string is called a square string if it can be obtained by concatenating two copies of the same string. For example, "abab", "aa" are square strings, while "aaa", "abba" are not. Hamming distanc…

Problem Description Given a set of sticks of various lengths, is it possible to join them end-to-end to form a square?   Input The first line of input contains N, the number of test cases. Each test case begins with an integer 4 <= M <= 20, the numb…

$dp$预处理,贪心. 因为$t$串前半部分和后半部分是一样的,所以只要构造前一半就可以了. 因为要求字典序最小,所以肯定是从第一位开始贪心选择,$a,b,c,d,...z$,一个一个尝试过去,如果发现某字符可行,那么该位就选择该字符. 第$i$位选择字符$X$可行的条件: 记这一位选择字符$X$的情况下,对$dis$的贡献为$Q$,$1$至$i-1$位对$dis$贡献和为$F$: 如果第$i+1$位至第$\frac{n}{2}$位,对$dis$的贡献可以凑出$m-Q-F$,那么该位选择$X$可…

题目链接:https://vjudge.net/contest/103424#problem/E 题目大意: 给你一堆硬币,让你分成两堆,分别给A,B两个人,求两人得到的最小差. 解题思路: 求解两人分得钱币的最小差值,巧妙地转化为01背包问题. sum代表这堆钱币的总价值,ans=sum/2,求出得钱较少的人的钱币总量,即在这堆钱币中挑选出一定量的钱币,使得它的总值为小于或等于ans的最大值,即将它转化为01背包问题,背包容量为ans,每一个钱币看成价值与体积相等的物品. #include <…