第四章 数组

4 AcWing 743 数组的行

a, b, c = int(input()),input(),[[float(x) for x in input().split()] for i in range(12)]
print('%.1lf'%(sum(c[a]) if b=='S' else sum(c[a])/12))
#分项定义语句
#基础数据读取与转换 列表类型使用及简写

5 AcWing 745 数组的右上半部分

a, b ,c= input(),[[float(y) for y in input().split()] for x in range(12)],0.
for i in range (12):
    for j in range(12):
        if (j>i):
            c+=b[i][j];
# c = sum([sum([i[x][y] for y in range(x+1,12)]) for x in range(12)])//该种写法也可以
print("%.1lf"%(c if a=='S' else c/((1+11)/2*11)))

6 AcWing 数组的左上半部分

a, b, ans = input(),[[float(x) for x in input().split()] for x in range(12)], 0.0
ans = sum([sum([b[x][y] for y in range(0,11-x)]) for x in range(12)])
print("%.1lf"%(ans if (a=='S') else ans/66))

7 AcWing 数组的上方区域

a, b, ans = input(),[[float(y) for y in input().split()] for i in range(12)], 0. 
for i in range(12):
    for j in range(12):
        if(j>i and i+j <11):
            ans += b[i][j]
# ans = sum([sum([b[i][j] for j in range(i+1,11-i)]) for i in range (5)])
# print("%.1lf"%(ans if a=='S' else ans/30))
print("%.1lf"%(ans if a=='S' else ans/30))

8 AcWing 数组的左方区域

a, b, c = input(), [[float(y) for y in input().split()] for i in range(12)] ,0.
#c = sum([sum([b[i][j] for i in range(j+1, 11-j)]) for j in range(5)])
for i in range(12):
    for j in range(12):
        if(i>j and i+j <11):
            c+=b[i][j]
print("%.1f"%(c if a=='S'else c/30))

9 763.平方矩阵

• C++版本

#include<iostream>
#include<algorithm>
using namespace std;
bool minmap(int a, int b)
{
    return a < b;
}
int main()
{
    int n;
    while(cin>>n){
        if(n==0) break;
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                // int d=min(min(i+1,j+1),min(n-i,n-j));
                int d = min({i+1,j+1,n-i,n-j});//min（）函数用法
                cout<<d<<" ";
            }
            cout<<endl;
        }
        cout<<endl;
       
    }
    
    return 0;
}

• Python版本

n=int(input())
while(n):
    for i in range(n):
        for j in range(n):
            print("%d "%(min(i+1, j+1,n-i,n-j)),end='')
        print()
    print()
    n = int(input())

10 740数组变换

#方法1 注意两种读取方式数据类型 方法2更优
a = [[int(y) for y in input().split()] for i in range(20)]
b = list(reversed(a))
for i in range(20):
    print("N[%d] = %d"%(i,b[i][0]))
#方法2 
a = list(reversed([int(input()) for x in range(20)]))
for i in range(20):
    print("N[%d] = %d"%(i,a[i]))

11 741 斐波那契数列

• 斐波那契通项公式（Fib）

#include<iostream>
using namespace std;
const int N = 100;
int main()
{
    int n;
    long int a[N]={0,1};
    for(int i=2;i<N;i++)
    {
        a[i]=a[i-1]+a[i-2];
    }
    cin>>n;
    while(n--)
    {
        int i;
        cin>>i;
        printf("Fib(%d) = %ld\n",i,a[i]);
    }
    return 0;
    
}

# 第一种方法
n = int(input())
b = [0,1]
for i in range(2,62):
    b.append(b[i-1]+b[i-2])
for i in range(n):
    a = int(input())
    print("Fib(%d) = %d"%(a,b[a]))
# 第二种方法
for x in range(int(input())):
    a, d = int(input()), 5**0.5
    print("Fib(%d) = %d"%(a,(1/d)*(((1+d)/2)**a-((1-d)/2)**a)))

//出现问题
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
    
    double c = sqrt(5);
    int n;
    cin>>n;
    while(n--)
    {
        int a;
        cin>>a;
        //cout<<(pow((1.0+c)/2.0,a)-pow((1.0-c)/2.0,a))*1.0/c<<endl;
        printf("Fib(%d) = %ld\n",a,int((pow((1.0+c)/2.0,a)-pow((1.0-c)/2.0,a))*1.0/c));
        //printf("Fib(%d) = %ld\n",a,(1.0/c)*((pow((1.0+c)/2.0,a)-pow((1.0-c)/2.0,a))));
    }
}


//正常
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
    
    double c = sqrt(5);
    int n;
    cin>>n;
    while(n--)
    {
        int a;
        cin>>a;
        long long int x =(pow((1.0+c)/2.0,a)-pow((1.0-c)/2.0,a))*1.0/c;
        //cout<<(pow((1.0+c)/2.0,a)-pow((1.0-c)/2.0,a))*1.0/c<<endl;
        printf("Fib(%d) = %ld\n",a,x);
        //printf("Fib(%d) = %ld\n",a,(1.0/c)*((pow((1.0+c)/2.0,a)-pow((1.0-c)/2.0,a))));
    }
}

742 最小数和它的位置

• 如何读取数据 并将单行数据转换成整数类型
• Python中 min 函数 index 函数的用法

n = input()
a = [int(x) for x in input().split()]
print("Minimum value: %d\nPosition: %d"%(min(a),a.index(min(a))))

• vector 容器插入元素做法、

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{
    vector<int> a;
    int n,b;
    cin>>n;
    for(int i =0;i<n;i++){
        cin>>b;
        a.push_back(b);
    }
    int min = *min_element(a.begin(),a.end());
    int index = min_element(a.begin(),a.end())-a.begin();
    printf("Minimum value: %d\nPosition: %d",min,index);
    return 0;

}

数组中的列

#include<iostream>
using namespace std;
const int N=12;
double a[N][N],sum=0;
int main()
{
    int n;
    char c;
    cin>>n>>c;
    for(int i=0;i<N;i++)
    {
        for(int j=0;j<N;j++){
            cin>>a[i][j];
        }
    }
    for(int j=0;j<N;j++){
        sum+=a[j][n];
    }
    if(c=='S')printf("%.1lf",sum);
    else printf("%.1lf",sum/12);
    return 0;
    
}

a,b,c,d=int(input()),input(),[[float(y) for y in input().split()] for i in range(12)],0.
x= sum([c[i][a] for i in range(12)])
print("%.1f"%(x if b=='S' else x/12))

数组的右下半部分

a,b=input(),[[float(y) for y in input().split()] for i in range(12)]
ans = sum([sum([b[i][j] for j in range(12-i,12)]) for i in range(12)])#(12-i,12)
print("%.1f"%(ans if a=='S' else ans/66))

数组的左下半部分

a,b =input(),[[float(y) for y in input().split()] for i in range(12)]
c = sum([sum([b[i][j] for j in range(i)]) for i in range(12)])
print("%.1lf" %(c if a=='S' else c/66))

数组的下方区域

数组的右方区域

平方矩阵2

a=int(input())
while(a):
    #print(a)
    a=int(input())
    for i in range(a):
        for j in range(a):
            print("%d"%(abs(i-j)+1),end=' ')
        print()
    print()
    
    
## 方法二 注意比较
while 1:
    n=int(input())
    if not n:
        break
    ans=[[0 for i in range(n)] for i in range(n)]
    i,j,num=0,0,1
    while i<n:
        ans[i][j]=num
        num+=1
        j+=1
        if j==n:
            j=i+1
            i+=1
            num=1
    for i in range(1,n):
        for j in range(i):
            ans[i][j]=ans[j][i]
    for i in range(n):
        print(*ans[i],sep=" ")
    print()

//方法一
#include<iostream>
using namespace std;
int main()
{
    int n;
    while(cin>>n,n!=0){
        int a[n][n]={0};
        for(int i=0;i<n;i++){
            for(int j=i;j<n;j++){
                a[i][j]=j-i+1;
            }
            for(int j=0;j<i;j++){
                a[i][j]=a[j][i];
            }
        }
        // for(int i=1;i<n;i++){
        //     for(int j=0;j<i;j++){
        //         a[i][j]=8;
        //     }
        // }
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                cout<<a[i][j]<<" ";
            }
            cout<<endl;
        }
        cout<<endl;
    }
}
//方法二
#include<iostream>
using namespace std;
int main(){
    int n;
    while(cin>>n,n){
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                cout<<abs(i-j)+1<<" ";
            }
            cout<<endl;
        }
        cout<<endl;
    }
    return 0;
}

平方矩阵3

• pow作用求幂 #include<cmath>

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;

int main(){
    int n,x;
    while(cin>>n,n)
    {
       for(int i=0;i<n;i++){
           for(int j=0;j<n;j++){
               x = pow(2,i+j);
               cout<<x<<" ";
           }
           cout<<endl;
       }
       cout<<endl;
    }
    return 0;
}

756 蛇形矩阵