A. Vasya the Hipster
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
One day Vasya the Hipster decided to count how many socks he had. It turned out that he had a red socks and b blue
socks.
According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.
Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.
Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Can you help him?
Input
The single line of the input contains two positive integers a and b (1 ≤ a, b ≤ 100)
— the number of red and blue socks that Vasya's got.
Output
Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day.
Sample test(s)
input
3 1
output
1 1
input
2 3
output
2 0
input
7 3
output
3 2
Note
In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day.
题目大意：
就是给定两个数m和n，然后求一下这两个数中的最小值，和最大值减完之后的一半，
太简单了，直接上代码：
/**
Author: ITAK
Date: 2015-09-29
**/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 10000+5;
const int mod = 1000000007;
const double eps = 1e-7;
int arr[maxn];
int main()
{
int m, n;
scanf("%d%d",&m, &n);
int ret = min(m, n);
int ans = max(m, n)-ret;
cout<<ret<<' '<<ans/2<<endl;
return 0;
}
B. Luxurious Houses
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were
built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all the houses with larger
numbers. In other words, a house is luxurious if the number of floors in it is strictly greater than in all the houses, which are located to the right from it. In this task it is assumed that the heights of floors in the houses are the same.
The new architect is interested in n questions, i-th
of them is about the following: "how many floors should be added to the i-th house to make it luxurious?" (for all i from 1 to n,
inclusive). You need to help him cope with this task.
Note that all these questions are independent from each other — the answer to the question for house i does not affect other answers
(i.e., the floors to the houses are not actually added).
Input
The first line of the input contains a single number n (1 ≤ n ≤ 105)
— the number of houses in the capital of Berland.
The second line contains n space-separated positive integers hi (1 ≤ hi ≤ 109),
where hi equals
the number of floors in the i-th house.
Output
Print n integers a1, a2, ..., an,
where number ai is
the number of floors that need to be added to the house number i to make it luxurious. If the house is already luxurious and nothing
needs to be added to it, then ai should
be equal to zero.
All houses are numbered from left to right, starting from one.
Sample test(s)
input
5 1 2 3 1 2
output
3 2 0 2 0
input
4 3 2 1 4
output
2 3 4 0
题目大意：
给定一个数m，然后有m个数arr[i],寻找能够大于右边所有数的最小值
解题思路：
首先找一个Max = 0,从右往左找，如果>Max,Max = arr[i],最小值是0;
否则最小值==Max - arr[i]+1;
上代码：
/**
Author: ITAK
Date: 2015-09-29
**/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 100000+5;
const int mod = 1000000007;
const double eps = 1e-7;
int arr[maxn], data[maxn];
int main()
{
int m;
scanf("%d",&m);
for(int i=0; i<m; i++)
scanf("%d",&arr[i]);
int Max = -1;
for(int i=m-1; i>=0; i--)
{
if(Max < arr[i])
{
Max = arr[i];
data[i] = 0;
}
else
data[i] = Max - arr[i]+1;
}
for(int i=0; i<m-1; i++)
cout<<data[i]<<" ";
cout<<data[m-1]<<endl;
return 0;
}
C. Developing Skills
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Petya loves computer games. Finally a game that he's been waiting for so long came out!
The main character of this game has n different skills, each of which is characterized by an integer ai from
0 to 100. The higher the number ai is,
the higher is the i-th skill of the character. The total rating of the character is calculated as the sum of the values of 
for
all i from 1 to n.
The expression ⌊ x⌋ denotes the result of rounding the number x down to
the nearest integer.
At the beginning of the game Petya got k improvement units as a bonus that he can use to increase the skills of his character and his
total rating. One improvement unit can increase any skill of Petya's character by exactly one. For example, if a4 = 46,
after using one imporvement unit to this skill, it becomes equal to 47. A hero's skill cannot rise higher more than 100. Thus, it is permissible that some of the units will remain unused.
Your task is to determine the optimal way of using the improvement units so as to maximize the overall rating of the character. It is not necessary to use all the improvement units.
Input
The first line of the input contains two positive integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 107)
— the number of skills of the character and the number of units of improvements at Petya's disposal.
The second line of the input contains a sequence of n integers ai (0 ≤ ai ≤ 100),
where ai characterizes
the level of the i-th skill of the character.
Output
The first line of the output should contain a single non-negative integer — the maximum total rating of the character that Petya can get using k or
less improvement units.
Sample test(s)
input
2 4 7 9
output
2
input
3 8 17 15 19
output
5
input
2 2 99 100
output
20
Note
In the first test case the optimal strategy is as follows. Petya has to improve the first skill to 10 by spending 3 improvement units, and the second skill to 10, by spending one improvement unit. Thus, Petya spends all his improvement units and the total rating
of the character becomes equal to lfloor frac{100}{10} rfloor + lfloor frac{100}{10} rfloor = 10 + 10 = 20.
In the second test the optimal strategy for Petya is to improve the first skill to 20 (by spending 3 improvement units) and to improve the third skill to 20 (in this case by spending 1 improvement units). Thus, Petya is left with 4 improvement units and he
will be able to increase the second skill to 19 (which does not change the overall rating, so Petya does not necessarily have to do it). Therefore, the highest possible total rating in this example is .
In the third test case the optimal strategy for Petya is to increase the first skill to 100 by spending 1 improvement unit. Thereafter, both skills of the character will be equal to 100, so Petya will not be able to spend the remaining improvement unit. So
the answer is equal to .
题目大意：
输入一个数，n和K，然后有n个数arr[i],让你求，在满足每个数<=100和k>0的情况下，每个值处以10取整的最大值，
注意arr[i]可以加不大于k的数，但是加完之后的值必须<=100，
解题思路：
就是先将每个值除以10之后取整的值记下来，然后再将它们对10取余之后的值记下来，再根据题意模拟判断就ok了，
但是一定要注意一些情况，arr[i]的值一直<=100
上代码：
/**
Author: ITAK
Date: 2015-09-29
**/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 100000+5;
const int mod = 1000000007;
const double eps = 1e-7;
int cmp(int a, int b)
{
return a > b;
}
int arr[maxn], data[maxn];
int main()
{
int n, k;
while(~scanf("%d%d",&n,&k))
{
int sum=0,K=0;
MM(data);
for(int i=0; i<n; i++)
{
scanf("%d",&arr[i]);
if(arr[i] != 100)
{
data[K++] = arr[i]%10;
sum += arr[i]/10;
}
else
sum += 10;
}
sort(data, data+K, cmp);
for(int i=0; i<K; i++)
{
k -= (10-data[i]);
data[i] =10;
if(k<0)
break;
if(data[i]%10==0)
sum ++;
}
if(n*10-sum > k/10)
sum += k/10;
else
sum = n*10;
cout<<sum<<endl;
}
return 0;
}