zoukankan      html  css  js  c++  java
  • cf div2 238 c

    C. Unusual Product
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.

    The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.

    Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:

    The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.

    However, there is much more to the homework. Chris has to process q queries; each query can be one of the following:

    1. given a row index i, flip all the values in the i-th row in A;
    2. given a column index i, flip all the values in the i-th column in A;
    3. find the unusual square of A.

    To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.

    Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?

    Input

    The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next nlines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th lineaij (0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.

    The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following:

    • i — flip the values of the i-th row;
    • i — flip the values of the i-th column;
    • 3 — output the unusual square of A.

    Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.

    Output

    Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.

    Sample test(s)
    input
    3
    1 1 1
    0 1 1
    1 0 0
    12
    3
    2 3
    3
    2 2
    2 2
    1 3
    3
    3
    1 2
    2 1
    1 1
    3
    output
    01001

    这个题目有个
     1 #include <cstdio>
     2 
     3 using namespace std;
     4 
     5 int n,p;
     6 
     7 int main() {
     8         scanf("%d",&n);
     9         int ans = 0;
    10         for(int i = 1; i <= n; ++i) {
    11                 for(int j = 1; j <= n; ++j) {
    12                         int v;
    13                         scanf("%d",&v);
    14                         if(i == j)
    15                         ans ^= v;
    16                 }
    17         }
    18 
    19         scanf("%d",&p);
    20         for(int i = 1; i <= p; ++i) {
    21                 int ch,v;
    22                 scanf("%d",&ch);
    23                 if(ch == 3) {
    24                         printf("%d",ans);
    25                 }
    26                 else {
    27                         scanf("%d",&v);
    28                         ans ^= 1;
    29                 }
    30         }
    31 
    32         return 0;
    33 
    34 }
    View Code
    规律,除了主对角线的元素自己与自己相乘,其他的元素相乘的组合每个都会出现两次,即他们的和必定是0所以只需要考虑主对角线的元素自行相乘的就可以了


  • 相关阅读:
    【276】◀▶ Python 字符串函数说明
    Spring事务配置的五种方式 巨全!不看后悔,一看必懂!
    Android Developers:两个视图渐变
    《Linux命令行与shell脚本编程大全》 第二十七章 学习笔记
    Android的TextView与Html相结合的用法
    嵌入式C语言优化小技巧
    vxworks获取系统时间编程
    【算法与数据结构】在n个数中取第k大的数(基础篇)
    字符集转换 字符类型转换 utf-8 gb2312 url
    java 从零开始,学习笔记之基础入门<Oracle_基础>(三十三)
  • 原文地址:https://www.cnblogs.com/hyxsolitude/p/3619884.html
Copyright © 2011-2022 走看看