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For a positive integer n let's define a function f:
f(n) = - 1 + 2 - 3 + .. + ( - 1)nn
Your task is to calculate f(n) for a given integer n.
The single line contains the positive integer n (1 ≤ n ≤ 1015).
Print f(n) in a single line.
4
2
5
-3
f(4) = - 1 + 2 - 3 + 4 = 2
f(5) = - 1 + 2 - 3 + 4 - 5 = - 3
题解：寒假归来先水几题再说……
这题主要是数学，分情况列公式就行了。
#include <iostream>
#include <cstring>
#include <cstdio>using namespace std;long long n;int main()
{scanf("%I64d",&n);if (!(n%2)) printf("%I64d",n/2);else printf("-%I64d",(n+1)/2);return 0;
}Let's define logical OR as an operation on two logical values (i. e. values that belong to the set {0, 1}) that is equal to 1 if either or both of the logical values is set to 1, otherwise it is 0. We can define logical OR of three or more logical values in the same manner:
where 
is equal to 1 if some ai = 1, otherwise it is equal to 0.
Nam has a matrix A consisting of m rows and n columns. The rows are numbered from 1 to m, columns are numbered from 1 to n. Element at row i (1 ≤ i ≤ m) and column j (1 ≤ j ≤ n) is denoted as Aij. All elements of A are either 0 or 1. From matrix A, Nam creates another matrix B of the same size using formula:
.
(Bij is OR of all elements in row i and column j of matrix A)
Nam gives you matrix B and challenges you to guess matrix A. Although Nam is smart, he could probably make a mistake while calculating matrix B, since size of A can be large.
The first line contains two integer m and n (1 ≤ m, n ≤ 100), number of rows and number of columns of matrices respectively.
The next m lines each contain n integers separated by spaces describing rows of matrix B (each element of B is either 0 or 1).
In the first line, print "NO" if Nam has made a mistake when calculating B, otherwise print "YES". If the first line is "YES", then also print mrows consisting of n integers representing matrix A that can produce given matrix B. If there are several solutions print any one.
2 2 1 0 0 0
NO
2 3 1 1 1 1 1 1
YES 1 1 1 1 1 1
2 3 0 1 0 1 1 1
YES 0 0 0 0 1 0
题解：这一题根据题目所给的情况构造，最后与原来的矩阵相比较就行了。
#include <iostream>
#include <cstring>
#include <cstdio>using namespace std;int m,n,a[105][105],b[105][105],c1[105],c2[105];int main()
{scanf("%d%d",&n,&m);for (int i=1;i<=n;i++)for (int j=1;j<=m;j++){scanf("%d",&b[i][j]);a[i][j]=1;}for (int i=1;i<=n;i++)for (int j=1;j<=m;j++) if (!b[i][j]){for (int k=1;k<=m;k++) a[i][k]=0;for (int k=1;k<=n;k++) a[k][j]=0;}for (int i=1;i<=n;i++)for (int j=1;j<=m;j++){int f=0;for (int k=1;k<=m;k++) f=f|a[i][k];for (int k=1;k<=n;k++) f=f|a[k][j];if (f!=b[i][j]){printf("NO\n");return 0;}}printf("YES\n");for (int i=1;i<=n;i++){for (int j=1;j<=m;j++) printf("%d ",a[i][j]);printf("\n");}return 0;
}Nam is playing with a string on his computer. The string consists of n lowercase English letters. It is meaningless, so Nam decided to make the string more beautiful, that is to make it be a palindrome by using 4 arrow keys: left, right, up, down.
There is a cursor pointing at some symbol of the string. Suppose that cursor is at position i (1 ≤ i ≤ n, the string uses 1-based indexing) now. Left and right arrow keys are used to move cursor around the string. The string is cyclic, that means that when Nam presses left arrow key, the cursor will move to position i - 1 if i > 1 or to the end of the string (i. e. position n) otherwise. The same holds when he presses the right arrow key (if i = n, the cursor appears at the beginning of the string).
When Nam presses up arrow key, the letter which the text cursor is pointing to will change to the next letter in English alphabet (assuming that alphabet is also cyclic, i. e. after 'z' follows 'a'). The same holds when he presses the down arrow key.
Initially, the text cursor is at position p.
Because Nam has a lot homework to do, he wants to complete this as fast as possible. Can you help him by calculating the minimum number of arrow keys presses to make the string to be a palindrome?
The first line contains two space-separated integers n (1 ≤ n ≤ 105) and p (1 ≤ p ≤ n), the length of Nam's string and the initial position of the text cursor.
The next line contains n lowercase characters of Nam's string.
Print the minimum number of presses needed to change string into a palindrome.
8 3 aeabcaez
6
A string is a palindrome if it reads the same forward or reversed.
In the sample test, initial Nam's string is: 
(cursor position is shown bold).
In optimal solution, Nam may do 6 following steps:
The result, 
, is now a palindrome.
题解：这一题首先可以看到只对一边处理就行，我们把P和要处理字符的全部放到左边处理。然后计算每一个字符处理的个数和移动的步数就行。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>using namespace std;char s[100005];
int n,p,l,ans,f[100005];int main()
{scanf("%d%d",&n,&p);scanf("%s",s);ans=0; memset(f,0,sizeof(f));for (int i=0;i<=n/2-1;i++){l=abs(s[n-i-1]-s[i]);if (l>13) l=26-l;ans+=l;if (l) f[i]=1;}if (p>n/2) p=n-p;else p--;int l1=0,l2=0;for (int i=0;i<=n/2-1;i++){if (i<=p && p-i>l1 && f[i]) l1=p-i;if (i>=p && i-p>l2 && f[i]) l2=i-p;}if (l1<=l2) ans=2*l1+l2+ans;else ans=ans+2*l2+l1;printf("%d",ans);return 0;
}
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