0.5 seconds
256 megabytes
standard input
standard output
Bomboslav likes to look out of the window in his room and watch lads outside playing famous shell game. The game is played by two persons: operator and player. Operator takes three similar opaque shells and places a ball beneath one of them. Then he shuffles the shells by swapping some pairs and the player has to guess the current position of the ball.
Bomboslav noticed that guys are not very inventive, so the operator always swaps the left shell with the middle one during odd moves (first, third, fifth, etc.) and always swaps the middle shell with the right one during even moves (second, fourth, etc.).
Let's number shells from 0 to 2 from left to right. Thus the left shell is assigned number 0, the middle shell is 1 and the right shell is 2. Bomboslav has missed the moment when the ball was placed beneath the shell, but he knows that exactly nmovements were made by the operator and the ball was under shell x at the end. Now he wonders, what was the initial position of the ball?
The first line of the input contains an integer n (1 ≤ n ≤ 2·109) — the number of movements made by the operator.
The second line contains a single integer x (0 ≤ x ≤ 2) — the index of the shell where the ball was found after n movements.
Print one integer from 0 to 2 — the index of the shell where the ball was initially placed.
4
2
1
1
1
0
In the first sample, the ball was initially placed beneath the middle shell and the operator completed four movements.
- During the first move operator swapped the left shell and the middle shell. The ball is now under the left shell.
- During the second move operator swapped the middle shell and the right one. The ball is still under the left shell.
- During the third move operator swapped the left shell and the middle shell again. The ball is again in the middle.
- Finally, the operators swapped the middle shell and the right shell. The ball is now beneath the right shell.
题意：三个位置上的东西 0 1 2 先0/1交换后1/2交换 告诉你交换n次之后的位置判断初始位置
题解：%6 水
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <cmath>
#include <map>
#define ll __int64
#define mod 1000000007
#define dazhi 2147483647
using namespace std;
int m[][];
int main()
{
m[][]=;
m[][]=;
m[][]=;
m[][]=;
m[][]=;
m[][]=; m[][]=;
m[][]=;
m[][]=;
m[][]=;
m[][]=;
m[][]=; m[][]=;
m[][]=;
m[][]=;
m[][]=;
m[][]=;
m[][]=;
ll n;
ll x;
scanf("%I64d",&n);
n=n%;
scanf("%I64d",&x);
printf("%d\n",m[x][n]);
return ;
}
2 seconds
256 megabytes
standard input
standard output
After the fourth season Sherlock and Moriary have realized the whole foolishness of the battle between them and decided to continue their competitions in peaceful game of Credit Cards.
Rules of this game are simple: each player bring his favourite n-digit credit card. Then both players name the digits written on their cards one by one. If two digits are not equal, then the player, whose digit is smaller gets a flick (knock in the forehead usually made with a forefinger) from the other player. For example, if n = 3, Sherlock's card is 123 and Moriarty's card has number 321, first Sherlock names 1 and Moriarty names 3 so Sherlock gets a flick. Then they both digit 2 so no one gets a flick. Finally, Sherlock names 3, while Moriarty names 1 and gets a flick.
Of course, Sherlock will play honestly naming digits one by one in the order they are given, while Moriary, as a true villain, plans to cheat. He is going to name his digits in some other order (however, he is not going to change the overall number of occurences of each digit). For example, in case above Moriarty could name 1, 2, 3 and get no flicks at all, or he can name 2, 3and 1 to give Sherlock two flicks.
Your goal is to find out the minimum possible number of flicks Moriarty will get (no one likes flicks) and the maximum possible number of flicks Sherlock can get from Moriarty. Note, that these two goals are different and the optimal result may be obtained by using different strategies.
The first line of the input contains a single integer n (1 ≤ n ≤ 1000) — the number of digits in the cards Sherlock and Moriarty are going to use.
The second line contains n digits — Sherlock's credit card number.
The third line contains n digits — Moriarty's credit card number.
First print the minimum possible number of flicks Moriarty will get. Then print the maximum possible number of flicks that Sherlock can get from Moriarty.
3
123
321
0
2
2
88
00
2
0
First sample is elaborated in the problem statement. In the second sample, there is no way Moriarty can avoid getting two flicks.
题意：两组n位数字a,b 对于相同的某一位i进行比较 a[i]<b[i] 则a的flicks值增加1 a[i]>b[i] 则b的flicks值增加1
任意排列两组的n位数字 问max(a_flicks) min(b_flicks)
题解：两种不同的贪心思路 具体看代码。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <cmath>
#include <map>
#define ll __int64
#define mod 1000000007
#define dazhi 2147483647
using namespace std;
char a[];
char b[];
int mp[][];
int main()
{
int n;
memset(mp,,sizeof(mp));
scanf("%d",&n);
getchar();
scanf("%s",a);
//cout<<a<<endl;
scanf("%s",b);
for(int i=;i<n;i++)
mp[][a[i]-'']++;
for(int i=;i<n;i++)
mp[][b[i]-'']++;
int exm1=;
int ans1=;
for(int i=;i<=;i++)
{
exm1+=mp[][i];
if(exm1>=mp[][i])
{
exm1-=mp[][i];
}
else
{
ans1+=(mp[][i]-exm1);
exm1=;
}
}
int exm2=mp[][];
int ans2=;
for(int i=;i>=;i--)
{
if(exm2>=mp[][i])
{
ans2+=mp[][i];
exm2=exm2-mp[][i];
}
else
{
ans2+=exm2;
exm2=;
}
exm2+=mp[][i];
}
printf("%d\n%d\n",ans1,ans2); return ;
}
1 second
256 megabytes
standard input
standard output
During the lesson small girl Alyona works with one famous spreadsheet computer program and learns how to edit tables.
Now she has a table filled with integers. The table consists of n rows and m columns. By ai, j we will denote the integer located at the i-th row and the j-th column. We say that the table is sorted in non-decreasing order in the column j if ai, j ≤ ai + 1, j for all i from 1 to n - 1.
Teacher gave Alyona k tasks. For each of the tasks two integers l and r are given and Alyona has to answer the following question: if one keeps the rows from l to r inclusive and deletes all others, will the table be sorted in non-decreasing order in at least one column? Formally, does there exist such j that ai, j ≤ ai + 1, j for all i from l to r - 1 inclusive.
Alyona is too small to deal with this task and asks you to help!
The first line of the input contains two positive integers n and m (1 ≤ n·m ≤ 100 000) — the number of rows and the number of columns in the table respectively. Note that your are given a constraint that bound the product of these two integers, i.e. the number of elements in the table.
Each of the following n lines contains m integers. The j-th integers in the i of these lines stands for ai, j (1 ≤ ai, j ≤ 109).
The next line of the input contains an integer k (1 ≤ k ≤ 100 000) — the number of task that teacher gave to Alyona.
The i-th of the next k lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n).
Print "Yes" to the i-th line of the output if the table consisting of rows from li to ri inclusive is sorted in non-decreasing order in at least one column. Otherwise, print "No".
5 4
1 2 3 5
3 1 3 2
4 5 2 3
5 5 3 2
4 4 3 4
6
1 1
2 5
4 5
3 5
1 3
1 5
Yes
No
Yes
Yes
Yes
No
In the sample, the whole table is not sorted in any column. However, rows 1–3 are sorted in column 1, while rows 4–5 are sorted in column 3.
题意：n*m的数字矩阵 k组查询[l,r] 问[l,r]行是否存在一列的值自上至下非递减
题解：ve[i]表示以第i行为末行最多有多少行满足题目要求。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <cmath>
#include <map>
#define ll __int64
#define mod 1000000007
#define dazhi 2147483647
using namespace std;
int n,m;
int mp[];
int ve[];
int re[];
int l,r;
int k;
int main()
{
memset(re,,sizeof(re));
scanf("%d %d",&n,&m);
for(int i=;i<n;i++)
{
for(int j=;j<m;j++)
scanf("%d",&mp[i*m+j]);
}
for(int i=;i<n*m;i++)
ve[i]=;
for(int j=;j<m;j++)
{
for(int i=;i<n;i++)
{
if(mp[i*m+j]>=mp[(i-)*m+j])
{
ve[i*m+j]+=ve[(i-)*m+j];
}
re[i]=max(re[i],ve[i*m+j]);
}
}
re[]=;
scanf("%d",&k);
for(int i=;i<=k;i++)
{
scanf("%d %d",&l,&r);
if(r--re[r-]+>l-)
printf("No\n");
else
printf("Yes\n");
}
return ;
}