A. Calculating Function
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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.
Input
The single line contains the positive integer n (1?≤?n?≤?1015).
Output
Print f(n) in a single line.
Sample test(s)
input
output
input
output
-3
Note
f(4)?=??-?1?+?2?-?3?+?4?=?2
f(5)?=??-?1?+?2?-?3?+?4?-?5?=??-?3
#include <iostream>#include <cstdio>#include <cstring>#include <stack>using namespace std;int main(){#ifdef xxz freopen("in.txt","r",stdin);#endif // xxz long long n; while(cin>>n) { long long sum = 0; sum += n/2; if(n%2) sum -= n; cout <p></p> <p class="sycode"> </p>
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<p class="sycode"> B. OR in Matrix </p> <p class="sycode"> </p>
<p class="sycode"> time limit per test </p> 1 second <p class="sycode"> </p>
<p class="sycode"> memory limit per test </p> 256 megabytes <p class="sycode"> </p>
<p class="sycode"> input </p> standard input <p class="sycode"> </p>
<p class="sycode"> output </p> standard output <p class="sycode"> </p>
<p> 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:</p> <p> where is equal to 1 if some ai?=?1, otherwise it is equal to 0.</p> <p> 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:</p> <p> .</p> <p> (Bij is OR of all elements in row i and column j of matrix A)</p> <p> 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.</p> <p class="sycode"> </p>
<p class="sycode"> Input </p> <p> The first line contains two integer m and n (1?≤?m,?n?≤?100), number of rows and number of columns of matrices respectively.</p> <p> 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).</p> <p class="sycode"> </p>
<p class="sycode"> Output </p> <p> 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.</p> <p class="sycode"> </p>
<p class="sycode"> Sample test(s) </p> <p class="sycode"> </p>
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<p class="sycode"> input </p> <pre style="代码" class="precsshei">2 21 00 0
output
NO
input
2 31 1 11 1 1
output
YES1 1 11 1 1
input
2 30 1 01 1 1
output
YES0 0 00 1 0
#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <sstream>#include <iomanip>#include <numeric>#include <cstring>#include <climits>#include <cassert>#include <complex>#include <cstdio>#include <string>#include <vector>#include <bitset>#include <queue>#include <stack>#include <cmath>#include <ctime>#include <list>#include <set>#include <map>using namespace std;int n , m;int g[1111][1111];int res[1111][1111];bool check_1(int row , int column){ int ok = 0; for(int k = 0 ; k <br> <p></p> <p class="sycode"> </p>
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<p class="sycode"> C. Palindrome Transformation </p> <p class="sycode"> </p>
<p class="sycode"> time limit per test </p> 1 second <p class="sycode"> </p>
<p class="sycode"> memory limit per test </p> 256 megabytes <p class="sycode"> </p>
<p class="sycode"> input </p> standard input <p class="sycode"> </p>
<p class="sycode"> output </p> standard output <p class="sycode"> </p>
<p> 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.</p> <p> 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).</p> <p> 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.</p> <p> Initially, the text cursor is at position p.</p> <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?</p> <p class="sycode"> </p>
<p class="sycode"> Input </p> <p> 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.</p> <p> The next line contains n lowercase characters of Nam's string.</p> <p class="sycode"> </p>
<p class="sycode"> Output </p> <p> Print the minimum number of presses needed to change string into a palindrome.</p> <p class="sycode"> </p>
<p class="sycode"> Sample test(s) </p> <p class="sycode"> </p>
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<p class="sycode"> input </p> <pre style="代码" class="precsshei">8 3aeabcaez
output
Note
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.
#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <sstream>#include <iomanip>#include <numeric>#include <cstring>#include <climits>#include <cassert>#include <complex>#include <cstdio>#include <string>#include <vector>#include <bitset>#include <queue>#include <stack>#include <cmath>#include <ctime>#include <list>#include <set>#include <map>using namespace std;int n , pos;char s[100100];int main(){#ifdef xxz freopen("in.txt","r",stdin);#endif int T,Case = 0;; while(cin>>n>>pos) { pos--; cin>>s; int len = strlen(s); int l = 0,r = n/2 - 1; if(n/2 =0 && s[r] == s[n-1-r] ) r--; int ans = 0; for(int i = l; i <br> <br> D: <p></p> <p> 树形DP，目前还不会，等打完广州区域赛在学习下树形DP再补上</p> </map></set></list></ctime></cmath></stack></queue></bitset></vector></string></cstdio></complex></cassert></climits></cstring></numeric></iomanip></sstream></fstream></iostream></algorithm></functional>
