2020/12/05第二次上机题目

最新推荐文章于 2021-12-08 14:29:07 发布

原创最新推荐文章于 2021-12-08 14:29:07 发布 · 221 阅读

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本文介绍了几种常见的矩阵操作和算法实现，包括交换矩阵的行、使用指针访问数组元素、判断字符串是否为回文串以及解决约瑟夫问题。通过具体的代码示例展示了这些操作的实现过程。

一、交换矩阵的行

#include <iostream>
using namespace std;
const int MAX_SIZE = 101;
int mat[MAX_SIZE][MAX_SIZE],n,m,max_sum,row;
int main() {
    m = 3,n=4;
    max_sum = -2147483648;
    for(int i = 1;i <= m;++i){
        for(int j = 1;j <= n;++j){
            mat[i][j] = i*j;
        }
    }
    for(int i = 1;i <= m;++i){
        for(int j = 1;j <= n;++j){
            mat[i][0] += mat[i][j];
        }
        if(max_sum < mat[i][0]){
	        max_sum = mat[i][0];
            row = i;
        }
    }
    swap(mat[1],mat[row]);
    for(int i = 1;i <= m;++i){
        for(int j = 1;j <= n;++j){
            cout<<mat[i][j]<<" ";
        }
        cout<<endl;
    }
    return 0;
}

二、指针访问数组

#include<iostream>
using namespace std;
const int ROW_SIZE = 101,COL_SIZE = 101;
int mat[ROW_SIZE][COL_SIZE],upper_sum,lower_sum,diognal_sum;
int (*pInt)[101];
int main(){
    for(int i = 0;i < ROW_SIZE;++i){
        for(int j = 0;j < COL_SIZE;++j){
            cin >> mat[i][j];
        }
    }
    for(auto row = mat;row < mat + ROW_SIZE;row++){
        for(auto col = *row + (row - mat);col < *row + COL_SIZE;++col){
            upper_sum += *col;
        }
    }
    for(auto row = mat;row < mat + ROW_SIZE;row++){
        for(auto col = *row ;col < *row  + (row - mat) + 1;++col){
            lower_sum += *col;
        }
    }
    for(auto row = mat;row < mat + ROW_SIZE;row++){
        auto col = *row + (row - mat);
        diognal_sum += *col;
    }
    return 0;
}

三、回文串

#include<iostream>
#include<cstring>
using namespace std;
string s;
int len;
bool Pald(int i,int j){
    return (s[i] == s[j])?((i == j || i + 1 == j)?true:Pald(i+1,j-1)):false;
}
int main(){
    char ch = getchar();
    while(ch != '\n'){
        if(('a' <= ch && ch <= 'z') || ('A' <= ch && ch <= 'Z')){
            if(ch >= 'A') s[len++] = ch - 'A' + 'a';
            else s[len++] = ch;
        }
        ch = getchar();
    }
    cout << (Pald(0,len - 1)?"Yes":"No");
    return 0;
}

四、约瑟夫问题

vector用吐了


#include<iostream>
#include<vector>
using namespace std;

vector<int> alive;
int n,m,ans;
int main(){
    cin >> n >> m;
    for(int i = 0;i < n;++i){
        alive.push_back(1);
    }
    vector<int>::iterator it = alive.begin();
    while (ans < n-1){
        int cnt = 1;
        while (cnt < m){
            if(it == alive.end()) it = alive.begin();
            else it++;
            while (*it != 1){
                if(it == alive.end()) it = alive.begin();
                else it++;
            }

            cnt++;
        }
        *it = 0;
        if(it == alive.end()) it = alive.begin();
        else it++;
        while (*it != 1){
            if(it == alive.end()) it = alive.begin();
            else it++;
        }
        ans++;
    }
    cout<<it - alive.begin()+1;
    return 0;
}

链表实现，一开始用vector遍历实在是要疯了

#include<iostream>
#include<vector>
using namespace std;

vector<int> alive[3];
int n,m,ans;
int main(){
    cin >> n >> m;
    for(int i = 0;i < n;++i){
        alive[0].push_back(1);
        alive[1].push_back(i+1);
        alive[2].push_back(i-1);
    }
    alive[1][n-1] = 0;
    alive[2][0] = n-1;
    int l = 0;
    while (ans < n-1){
        int cnt = 1;
        while (cnt < m){
            l = alive[1][l];
            cnt++;
        }
        alive[1][alive[2][l]] = alive[1][l];
        alive[2][alive[1][l]] = alive[2][l];
        alive[0][l] = 0;
        l = alive[1][l];
        ans++;
    }
    for(int i = 0;i < n;++i){
        if(alive[0][i]) cout<<i+1;
    }
    return 0;
}