AtCoder Beginner Contest 072
今天打得真的很爽,15分钟就AK了,第一次在13名。。。在过生日时打真的是我的幸运日,哈哈哈。
Time limit : 2sec / Memory limit : 256MB
Score : 100 points
We have a sandglass that runs for X seconds. The sand drops from the upper bulb at a rate of 1 gram per second. That is, the upper bulb initially contains X grams of sand.
How many grams of sand will the upper bulb contains after t seconds?
The input is given from Standard Input in the following format:
X t
Print the number of sand in the upper bulb after t second.
100 17
83
17 out of the initial 100 grams of sand will be consumed, resulting in 83 grams.
有x克,每秒减少一克,问t秒后还有多少克,水题。
1 #include <bits/stdc++.h> 2 #define ll long long 3 using namespace std; 4 5 int main() { 6 int x, t; 7 cin >> x >> t; 8 printf("%d\n",max(0,x-t)); 9 return 0; 10 }
Time limit : 2sec / Memory limit : 256MB
Score : 200 points
You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.
The input is given from Standard Input in the following format:
s
Print the string obtained by concatenating all the characters in the odd-numbered positions.
atcoder
acdr
Extract the first character a
, the third character c
, the fifth character d
and the seventh character r
to obtain acdr
.
把位数为奇数的输入下,水题。
1 #include <bits/stdc++.h> 2 #define ll long long 3 using namespace std; 4 const int N = 1e5+10; 5 char str[N]; 6 int main() { 7 cin >> str; 8 for(int i = 0; str[i]; i++) { 9 if(i %2 == 0) printf("%c",str[i]); 10 } 11 printf("\n"); 12 return 0; 13 }
Time limit : 2sec / Memory limit : 256MB
Score : 300 points
You are given an integer sequence of length N, a1,a2,…,aN.
For each 1≤i≤N, you have three choices: add 1 to ai, subtract 1 from ai or do nothing.
After these operations, you select an integer X and count the number of i such that ai=X.
Maximize this count by making optimal choices.
The input is given from Standard Input in the following format:
N a1 a2 .. aN
Print the maximum possible number of i such that ai=X.
7 3 1 4 1 5 9 2
4
For example, turn the sequence into 2,2,3,2,6,9,2 and select X=2 to obtain 4, the maximum possible count.
每位可以加减一或不动,求相同数字的个数最大值。水题。
1 #include <bits/stdc++.h> 2 #define ll long long 3 using namespace std; 4 const int N = 1e5+10; 5 // int a[N]; 6 map<int, int> mp; 7 int main() { 8 int n, x; 9 scanf("%d",&n); 10 for(int i = 0; i < n; i ++) { 11 scanf("%d",&x); 12 mp[x]++; 13 mp[x+1]++; 14 mp[x-1]++; 15 } 16 map<int, int> ::iterator it; 17 int MAX = 1; 18 for(it = mp.begin(); it != mp.end(); ++it) { 19 if(MAX < (*it).second){ 20 MAX = (*it).second; 21 } 22 } 23 printf("%d\n",MAX); 24 return 0; 25 }
Time limit : 2sec / Memory limit : 256MB
Score : 400 points
You are given a permutation p1,p2,…,pN consisting of 1,2,..,N. You can perform the following operation any number of times (possibly zero):
Operation: Swap two adjacent elements in the permutation.
You want to have pi≠i for all 1≤i≤N. Find the minimum required number of operations to achieve this.
The input is given from Standard Input in the following format:
N p1 p2 .. pN
Print the minimum required number of operations
5 1 4 3 5 2
2
Swap 1 and 4, then swap 1 and 3. p is now 4,3,1,5,2 and satisfies the condition. This is the minimum possible number, so the answer is 2.
每次可以和相邻的交换位置,求最少的交换次数使的a[i] != i,继续水题。
1 #include <bits/stdc++.h> 2 #define ll long long 3 using namespace std; 4 const int N = 1e5+10; 5 int a[N]; 6 int main() { 7 int n; 8 scanf("%d",&n); 9 for(int i = 1; i <= n; i ++) { 10 scanf("%d", &a[i]); 11 } 12 int ans = 0; 13 for(int i = 1; i < n; i ++) { 14 if(a[i] == i) { 15 swap(a[i], a[i+1]); 16 ans++; 17 } 18 } 19 if(a[n] == n) ans++; 20 printf("%d\n",ans); 21 return 0; 22 }优质内容筛选与推荐>>