世上只有一种英雄主义，就是在认清生活真相之后依然热爱生活。

大家有空去我的网站逛逛呀！ http://www.mxnzqh.cn/

A - Max Sum Plus Plus

Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem.

Given a consecutive number sequence S 1, S 2, S3, S 4 ... S x, ... S n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S x ≤ 32767). We define a function sum(i, j) = S i + ... + S j (1 ≤ i ≤ j ≤ n).

Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i 1, j 1) + sum(i 2, j 2) + sum(i 3, j 3) + ... + sum(i m, jm) maximal (i x ≤ i y ≤ j x or i x ≤ j y ≤ j xis not allowed).

But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(i x, j x)(1 ≤ x ≤ m) instead. ^_^

Input

Each test case will begin with two integers m and n, followed by n integers S 1, S 2, S 3 ... S n.
Process to the end of file.

Output

Output the maximal summation described above in one line.

Sample Input

1 3 1 2 3
2 6 -1 4 -2 3 -2 3

Sample Output

6
8

Hint

Huge input, scanf and dynamic programming is recommended.

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <vector>
#include <ctime>
#include <cctype>
#include <bitset>
#include <utility>
#include <sstream>
#include <complex>
#include <iomanip>
#define inf 0x3f3f3f3f
typedef long long ll;
using namespace std;
//dp[i][j]表示前j个数选择i段的最大值
int m,n,s[1000010],dp[1000010],q[1000010],mx;
int main()
{while(cin>>m>>n){for(int i=1; i<=n; i++)cin>>s[i];memset(dp,0,sizeof(dp));memset(q,0,sizeof(q));for(int i=1; i<=m; i++){mx=-inf;for(int j=i; j<=n; j++){dp[j]=max(dp[j-1],q[j-1])+s[j];q[j-1]=mx;mx=max(dp[j],mx);}}cout<<mx<<endl;}return 0;
}

B - Ignatius and the Princess IV

"OK, you are not too bad, em... But you can never pass the next test." feng5166 says.

"I will tell you an odd number N, and then N integers. There will be a special integer among them, you have to tell me which integer is the special one after I tell you all the integers." feng5166 says.

"But what is the characteristic of the special integer?" Ignatius asks.

"The integer will appear at least (N+1)/2 times. If you can't find the right integer, I will kill the Princess, and you will be my dinner, too. Hahahaha....." feng5166 says.

Can you find the special integer for Ignatius?

Input

The input contains several test cases. Each test case contains two lines. The first line consists of an odd integer N(1<=N<=999999) which indicate the number of the integers feng5166 will tell our hero. The second line contains the N integers. The input is terminated by the end of file.

Output

For each test case, you have to output only one line which contains the special number you have found.

Sample Input

5
1 3 2 3 3
11
1 1 1 1 1 5 5 5 5 5 5
7
1 1 1 1 1 1 1

Sample Output

3
5
1

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <vector>
#include <ctime>
#include <cctype>
#include <bitset>
#include <utility>
#include <sstream>
#include <complex>
#include <iomanip>
#define inf 0x3f3f3f3f
typedef long long ll;
using namespace std;
int N,s[1000010],x,jg;
int main()
{while(cin>>N){memset(s,0,sizeof(s));for(int i=0; i<N; i++){cin>>x;s[x]++;if(s[x]==N/2)jg=x;}cout<<jg<<endl;}return 0;
}

C - Monkey and Banana

A group of researchers are designing an experiment to test the IQ of a monkey. They will hang a banana at the roof of a building, and at the mean time, provide the monkey with some blocks. If the monkey is clever enough, it shall be able to reach the banana by placing one block on the top another to build a tower and climb up to get its favorite food.

The researchers have n types of blocks, and an unlimited supply of blocks of each type. Each type-i block was a rectangular solid with linear dimensions (xi, yi, zi). A block could be reoriented so that any two of its three dimensions determined the dimensions of the base and the other dimension was the height.

They want to make sure that the tallest tower possible by stacking blocks can reach the roof. The problem is that, in building a tower, one block could only be placed on top of another block as long as the two base dimensions of the upper block were both strictly smaller than the corresponding base dimensions of the lower block because there has to be some space for the monkey to step on. This meant, for example, that blocks oriented to have equal-sized bases couldn't be stacked.

Your job is to write a program that determines the height of the tallest tower the monkey can build with a given set of blocks.

Input

The input file will contain one or more test cases. The first line of each test case contains an integer n,
representing the number of different blocks in the following data set. The maximum value for n is 30.
Each of the next n lines contains three integers representing the values xi, yi and zi.
Input is terminated by a value of zero (0) for n.

Output

For each test case, print one line containing the case number (they are numbered sequentially starting from 1) and the height of the tallest possible tower in the format "Case case: maximum height = height".

Sample Input

Sample Output

Case 1: maximum height = 40
Case 2: maximum height = 21
Case 3: maximum height = 28
Case 4: maximum height = 342

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <vector>
#include <ctime>
#include <cctype>
#include <bitset>
#include <utility>
#include <sstream>
#include <complex>
#include <iomanip>
#define inf 0x3f3f3f3f
typedef long long ll;
using namespace std;
int n,dp[210],tt=1,ct;
struct node
{int x;int y;int z;
} p[210];
bool cmp(node a,node b)
{if(a.x==b.x)return a.y>b.y;return a.x>b.x;
}
void plzh(int xx,int yy,int zz)
{p[ct].x=xx;p[ct].y=yy;p[ct].z=zz;ct++;
}
int main()
{while(scanf("%d",&n)&&n!=0){int xx,yy,zz,jg=0;ct=0;memset(dp,0,sizeof(dp));for(int i=0; i<n; i++){scanf("%d%d%d",&xx,&yy,&zz);plzh(xx,yy,zz);plzh(xx,zz,yy);plzh(yy,xx,zz);plzh(yy,zz,xx);plzh(zz,xx,yy);plzh(zz,yy,xx);}sort(p,p+ct,cmp);for(int i=0; i<ct; i++){dp[i]=p[i].z;for(int j=0; j<i; j++){if(p[i].x<p[j].x&&p[i].y<p[j].y)dp[i]=max(dp[i],dp[j]+p[i].z);jg=max(jg,dp[i]);}}printf("Case %d: maximum height = %d\n",tt++,jg);}return 0;
}

D - Doing Homework

Ignatius has just come back school from the 30th ACM/ICPC. Now he has a lot of homework to do. Every teacher gives him a deadline of handing in the homework. If Ignatius hands in the homework after the deadline, the teacher will reduce his score of the final test, 1 day for 1 point. And as you know, doing homework always takes a long time. So Ignatius wants you to help him to arrange the order of doing homework to minimize the reduced score.

Input

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case start with a positive integer N(1<=N<=15) which indicate the number of homework. Then N lines follow. Each line contains a string S(the subject's name, each string will at most has 100 characters) and two integers D(the deadline of the subject), C(how many days will it take Ignatius to finish this subject's homework).

Note: All the subject names are given in the alphabet increasing order. So you may process the problem much easier.

Output

For each test case, you should output the smallest total reduced score, then give out the order of the subjects, one subject in a line. If there are more than one orders, you should output the alphabet smallest one.

Sample Input

2
3
Computer 3 3
English 20 1
Math 3 2
3
Computer 3 3
English 6 3
Math 6 3

Sample Output

2
Computer
Math
English
3
Computer
English
Math

Hint

In the second test case, both Computer->English->Math and Computer->Math->English leads to reduce 3 points, but the
word "English" appears earlier than the word "Math", so we choose the first order. That is so-called alphabet order.

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <vector>
#include <ctime>
#include <cctype>
#include <bitset>
#include <utility>
#include <sstream>
#include <complex>
#include <iomanip>
#define inf 0x3f3f3f3f
typedef long long ll;
using namespace std;
int T,N,yx[40010],sj[40010],dp[40010];
struct node
{char s[110];int D;int C;
} p[40010];
void shuchu(int m)
{if(!m)return;shuchu(m-(1<<yx[m]));cout<<p[yx[m]].s<<endl;
}
int main()
{cin>>T;while(T--){cin>>N;getchar();for(int i=0; i<N; i++)cin>>p[i].s>>p[i].D>>p[i].C;int m=1<<N;for(int i=1; i<m; i++){dp[i]=inf;for(int j=N-1; j>=0; j--){int ls=1<<j;if(!(i&ls))continue;int fs=sj[i-ls]+p[j].C-p[j].D;if(fs<0)fs=0;if(dp[i]>dp[i-ls]+fs){dp[i]=dp[i-ls]+fs;sj[i]=sj[i-ls]+p[j].C;yx[i]=j;}}}cout<<dp[m-1]<<endl;shuchu(m-1);}return 0;
}

E - Super Jumping! Jumping! Jumping!

Nowadays, a kind of chess game called “Super Jumping! Jumping! Jumping!” is very popular in HDU. Maybe you are a good boy, and know little about this game, so I introduce it to you now.

The game can be played by two or more than two players. It consists of a chessboard（棋盘）and some chessmen（棋子）, and all chessmen are marked by a positive integer or “start” or “end”. The player starts from start-point and must jumps into end-point finally. In the course of jumping, the player will visit the chessmen in the path, but everyone must jumps from one chessman to another absolutely bigger (you can assume start-point is a minimum and end-point is a maximum.). And all players cannot go backwards. One jumping can go from a chessman to next, also can go across many chessmen, and even you can straightly get to end-point from start-point. Of course you get zero point in this situation. A player is a winner if and only if he can get a bigger score according to his jumping solution. Note that your score comes from the sum of value on the chessmen in you jumping path.
Your task is to output the maximum value according to the given chessmen list.

Input

Input contains multiple test cases. Each test case is described in a line as follow:
N value_1 value_2 …value_N
It is guarantied that N is not more than 1000 and all value_i are in the range of 32-int.
A test case starting with 0 terminates the input and this test case is not to be processed.

Output

For each case, print the maximum according to rules, and one line one case.

Sample Input

Sample Output

4
10
3

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <vector>
#include <ctime>
#include <cctype>
#include <bitset>
#include <utility>
#include <sstream>
#include <complex>
#include <iomanip>
#define inf 0x3f3f3f3f
typedef long long ll;
using namespace std;
int n,dp[1010], a[1010];
int main()
{while(cin>>n&&n!=0){int jg = 0;for(int i=0; i<n; i++){cin>>a[i];dp[i] = a[i];for(int j=0; j<i; j++)if(a[j] < a[i])dp[i] = max(dp[i], dp[j]+a[i]);jg = max(jg, dp[i]);}cout<<jg<<endl;}return 0;
}

F - Piggy-Bank

Before ACM can do anything, a budget must be prepared and the necessary financial support obtained. The main income for this action comes from Irreversibly Bound Money (IBM). The idea behind is simple. Whenever some ACM member has any small money, he takes all the coins and throws them into a piggy-bank. You know that this process is irreversible, the coins cannot be removed without breaking the pig. After a sufficiently long time, there should be enough cash in the piggy-bank to pay everything that needs to be paid.

But there is a big problem with piggy-banks. It is not possible to determine how much money is inside. So we might break the pig into pieces only to find out that there is not enough money. Clearly, we want to avoid this unpleasant situation. The only possibility is to weigh the piggy-bank and try to guess how many coins are inside. Assume that we are able to determine the weight of the pig exactly and that we know the weights of all coins of a given currency. Then there is some minimum amount of money in the piggy-bank that we can guarantee. Your task is to find out this worst case and determine the minimum amount of cash inside the piggy-bank. We need your help. No more prematurely broken pigs!

Input

The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing two integers E and F. They indicate the weight of an empty pig and of the pig filled with coins. Both weights are given in grams. No pig will weigh more than 10 kg, that means 1 <= E <= F <= 10000. On the second line of each test case, there is an integer number N (1 <= N <= 500) that gives the number of various coins used in the given currency. Following this are exactly N lines, each specifying one coin type. These lines contain two integers each, Pand W (1 <= P <= 50000, 1 <= W <=10000). P is the value of the coin in monetary units, W is it's weight in grams.

Output

Print exactly one line of output for each test case. The line must contain the sentence "The minimum amount of money in the piggy-bank is X." where X is the minimum amount of money that can be achieved using coins with the given total weight. If the weight cannot be reached exactly, print a line "This is impossible.".

Sample Input

Sample Output

The minimum amount of money in the piggy-bank is 60.
The minimum amount of money in the piggy-bank is 100.
This is impossible.

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <vector>
#include <ctime>
#include <cctype>
#include <bitset>
#include <utility>
#include <sstream>
#include <complex>
#include <iomanip>
#define inf 0x3f3f3f3f
typedef long long ll;
using namespace std;
//完全背包
int n,t,e,f,w[50010],p[50010],dp[50010];
int main()
{cin>>t;while(t--){cin>>e>>f;f-=e;//硬币重量cin>>n;for(int i=0; i<n; i++)cin>>p[i]>>w[i];for(int i=0; i<=f; i++)dp[i]=inf;dp[0]=0;for(int i=0; i<n; i++)for(int j=w[i]; j<=f; j++)dp[j]=min(dp[j],dp[j-w[i]]+p[i]);if(dp[f]==inf)printf("This is impossible.\n");elseprintf("The minimum amount of money in the piggy-bank is %d.\n",dp[f]);}return 0;
}

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YTU2018级每周训练-动态规划1

世上只有一种英雄主义，就是在认清生活真相之后依然热爱生活。

大家有空去我的网站逛逛呀！ http://www.mxnzqh.cn/

A - Max Sum Plus Plus

B - Ignatius and the Princess IV

C - Monkey and Banana

D - Doing Homework

E - Super Jumping! Jumping! Jumping!

F - Piggy-Bank

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