Beans Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 2845 Description Bean-eating is an interesting game, everyone owns an M*N matrix, which is filled with different qualities beans. Meantime,…

Beans Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3521    Accepted Submission(s): 1681 Problem Description Bean-eating is an interesting game, everyone owns an M*N matrix, which is filled w…

Beans Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2596    Accepted Submission(s): 1279 Problem Description Bean-eating is an interesting game, everyone owns an M*N matrix, which is filled w…

Problem Description Bean-eating is an interesting game, everyone owns an M*N matrix, which is filled with different qualities beans. Meantime, there is only one bean in any 1*1 grid. Now you want to eat the beans and collect the qualities, but everyo…

Beans Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2518    Accepted Submission(s): 1250 Problem Description Bean-eating is an interesting game, everyone owns an M*N matrix, which is filled wi…

Problem Description Bean-eating is an interesting game, everyone owns an M*N matrix, which is filled with different qualities beans. Meantime, there is only one bean in any 1*1 grid. Now you want to eat the beans and collect the qualities, but everyo…

Problem Description Bean-eating is an interesting game, everyone owns an M*N matrix, which is filled with different qualities beans. Meantime, there is only one bean in any 1*1 grid. Now you want to eat the beans and collect the qualities, but everyo…

Beans Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4456    Accepted Submission(s): 2105 Problem Description Bean-eating is an interesting game, everyone owns an M*N matrix, which is filled w…

Beans Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3418    Accepted Submission(s): 1629 Problem Description Bean-eating is an interesting game, everyone owns an M*N matrix, which is filled wi…

Saving Beans Saving Beans Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5761    Accepted Submission(s): 2310 Problem Description Although winter is far away, squirrels have to work day and nig…

题意:容易理解. 分析:以后碰到这种类型的题,就要考虑把矩阵先按行来处理,再按列处理.先算出每行能够能够得到的最大值,然后按列处理即可. 代码实现: #include<stdio.h> #include<string.h> int n,m; ],dp[][],b[]; int max(int x,int y) { return x>y?x:y; } int main() { int i,j,res; while(scanf("%d%d",&n,&a…

题意:给你n袋豆子,每袋都有w[i]个豆子,接着任选连续任意个袋子的豆子合在一起放入容量为p的多个袋子里(每个袋子必须放满),问剩余的豆子数<=k时,能放满最多的袋子的个数 题解:个数与p都比较大,直接模拟O(n^2),余数处理(dp)O(n*p)都会超时. 我们可以首先抽象出一个公式来:设前缀和为sum[i],则我们求(j+1,i)的豆子的数量时会使用sum[i]-sum[j]来求,而我们需要求的就是max(sum[i]-sum[j]),条件的是 (sum[i]-sum[j])%P<=k —…

/*递推公式dp[i]=MAX(dp[i-1],dp[i-2]+a[j])*/ #include<stdio.h> #include<string.h> #define N 210000 int a[N],f[N],dp[N]; int Max(int v,int vv) { return v>vv?v:vv; } int main() { int n,m,i,j,k; while(scanf("%d%d",&n,&m)!=EOF) { m…

#include<stdio.h> #define N 200100 int f[N]; int  a[N],n; int main() { int m,j,i,suma,sumb,sumc,sumd; while(scanf("%d%d",&m,&n)!=EOF) {        for(i=1;i<=m;i++)   for(j=1;j<=n;j++)   scanf("%d",&f[(i-1)*n+j]);  …

DP 158:11:22 1205:00:00   Overview Problem Status Rank (56) Discuss Current Time: 2015-11-26 19:11:23 Contest Type: Private Start Time: 2015-11-20 05:00:00 Contest Status: Running End Time: 2016-01-09 10:00:00 Manager: qwerqqq Clone this contest Edit…

hdu 3037 Saving Beans 题目大意:n个数,和不大于m的情况,结果模掉p,p保证为素数. 解题思路:隔板法,C(nn+m)多选的一块保证了n个数的和小于等于m.可是n,m非常大,所以用到Lucas定理. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef long long ll; ll n, m, p; ll qPow (ll a…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=2845 题意:吃豆子游戏 , 当你吃了一个格子的豆子 , 该格子左右两个和上下两行就不能吃了 , 输入每个格子的豆子数 , 求你最多能吃多少颗豆子? 在最大连续数列的基础上,改变了,求最大不连续和? 我们可以先求单独的每一行的的最大不连续和,相当于对矩阵进行压缩,将n列压缩成了一列.然后再求,这一列的最大不连续和. dp方程: m[j]=max(m[j-1],m[j-2]+temp) 对temp两种选择…

Saving Beans Time Limit: 3000ms Memory Limit: 32768KB This problem will be judged on HDU. Original ID: 303764-bit integer IO format: %I64d      Java class name: Main   Although winter is far away, squirrels have to work day and night to save beans. T…

http://acm.hdu.edu.cn/showproblem.php?pid=3037 Lucas定理模板. 现在才写,noip滚粗前兆QAQ #include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef long long ll; int jc[100003]; int p; int ipow(int x, int b) { ll t = 1, w = x;…

Saving Beans Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4315    Accepted Submission(s): 1687 Problem Description Although winter is far away, squirrels have to work day and night to save be…

Problem Description Although winter is far away, squirrels have to work day and night to save beans. They need plenty of food to get through those long cold days. After some time the squirrel family thinks that they have to solve a problem. They supp…

Saving Beans Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2079    Accepted Submission(s): 748 Problem Description Although winter is far away, squirrels have to work day and night to save bea…

主题链接:pid=3037">http://acm.hdu.edu.cn/showproblem.php?pid=3037 推出公式为C(n + m, m) % p. 用Lucas定理求大组合数取模的值 代码: #include <stdio.h> #include <string.h> #include <algorithm> using namespace std; int t; long long n, m, p; long long pow(lo…

Saving Beans Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8258    Accepted Submission(s): 3302 Problem Description Although winter is far away, squirrels have to work day and night to save be…

Problem Description Although winter is far away, squirrels have to work day and night to save beans. They need plenty of food to get through those long cold days. After some time the squirrel family thinks that they have to solve a problem. They supp…

Saving Beans Time Limit: 3000 MS Memory Limit: 32768 K Problem Description Although winter is far away, squirrels have to work day and night to save beans. They need plenty of food to get through those long cold days. After some time the squirrel fam…

解题思路: 直接求C(n+m , m) % p , 由于n , m ,p都非常大,所以要用Lucas定理来解决大组合数取模的问题. #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string&g…

意甲冠军: 1. 对于每一行是,对不相邻的同一时间数取: 2.它是相同的列,相邻行不能同时服用: 3.因此,我们可以得到状态方程:dp[i]=dp[i-1]>(dp[i-2]+a[i])?dp[i-1]:dp[i-2]+a[i].先对每一行运用,在对每一行求出的和作为一组运用.可得终于结果. #include<iostream> using namespace std; int col[200001]; int dp[200001]; int GetMaxRow(int a[],int…

#include<cstdio> #include<cstdlib> #include<algorithm> #include<cstring> #include<vector> #define MAXN 100000+10 #define ll long long #define pb push_back #define ft first #define sc second #define mp make_pair #define pil pa…

题意:问用不超过 m 颗种子放到 n 棵树中,有多少种方法. 析:题意可以转化为 x1 + x2 + .. + xn = m,有多少种解,然后运用组合的知识就能得到答案就是 C(n+m, m). 然后就求这个值,直接求肯定不好求,所以我们可以运用Lucas定理,来分解这个组合数,也就是Lucas(n,m,p)=C(n%p,m%p)* Lucas(n/p,m/p,p). 然后再根据费马小定理就能做了. 代码如下: 第一种: #pragma comment(linker, "/STACK:10240…