软盟2练习答案

1.签到题I

这题排个序输出第k个下标就好，注意输入如果从0开始就是输出k-1

c++：

#include<iostream>

#include<algorithm>

using namespace std;

const int N = 100010;

int arr[N];

int n, k;

int main()

{

cin >> n >> k;

for (int i = 0; i < n; i++) cin >> arr[i];

sort(arr, arr + n);

cout << arr[k-1] << endl;

return 0;

}

2.A题

这题算法涉及到了快速幂，灵活运用快速幂即可

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
int qmi(int a, int b, int p)
{
    int res = 1 % p;
    while (b)
    {
        if (b & 1) res = res * (LL)a % p;
        a = a * (LL)a % p;
        b >>= 1;
    }
    return res;
}
 
int main()
{
    int n, m, k, x;
    scanf("%d%d%d%d", &n, &m, &k, &x);
 
    printf("%d\n", (x + qmi(10, k, n) * (LL)m) % n);
    return 0;
}

3.B题

本来想考区间和或差分的，没发现这题要求边修改边查询，就得用树状数组或者线段树做，这边推荐用树状数组的知识点做更简单

#include <iostream>

using namespace std;
typedef long long LL;
// 树状数组中最重要的操作，修改和查询

const int N = 1000010;

int a[N];
LL tr[N];

int n , q;

int lowbit(int x) {
	return x & (-x);
}

void add(int u , int x) {
	for (int i = u ; i <= n ; i += lowbit(i)) tr[i] += x;
}

LL sum(int x) {
	LL res = 0;
	for (int i = x ; i ; i -= lowbit(i)) res += tr[i];
	return res;
}

int main() {
	scanf("%d%d", &n, &q);
	for (int i = 1 ; i <= n ; i ++) scanf("%d", &a[i]);
	for (int i = 1 ; i <= n ; i ++) add(i , a[i]); // 建立树状数组 
	for (int i = 1 ; i <= q ; i ++) {
		int c;
		scanf("%d", &c);
		if (c == 1) {
			int u , x;
			scanf("%d%d", &u, &x);
			add(u , x);
		} else {
			int l , r;
			scanf("%d%d", &l, &r);
			printf("%lld\n", sum(r) - sum(l - 1));
		}
	}

	return 0;
}

第四题：小红的扫雷游戏

这题很考验思维，相当于怎么用用代码编写我们平时玩的扫雷游戏的推断思想，非常灵活这里提供用dfs的做法

#include<iostream>
#include<algorithm>
using namespace std;
const int N=5;
char g[N][N];
int cnt[N][N];
int d;
int dx[8]={-1,-1,-1,0,0,1,1,1};
int dy[8]={-1,0,1,1,-1,-1,0,1};
  
void dfs(int x,int y)
{
    if(y==4)
    {
        y=0;x++;
    }
    if(x==4)
    {
        for(int i=0;i<4;i++)
            for(int j=0;j<4;j++)
            {
                if(g[i][j]>='0'&&g[i][j]<='9')
                {
                    int k=g[i][j]-'0';
                    int t=0;
                    for(int u=0;u<8;u++)
                    {
                        int nx=i+dx[u];int ny=j+dy[u];
                        if(nx<0||nx>=8||ny<0||ny>=8)continue;
                        if(g[nx][ny]=='X')t++;
                    }
                    if(t!=k)return;
                }
            }
        d++;
        for(int i=0;i<4;i++)
            for(int j=0;j<4;j++)
                if(g[i][j]=='X')cnt[i][j]++;
        return;
    }
    if(g[x][y]>='0'&&g[x][y]<='9')dfs(x,y+1);
    else
    {
        g[x][y]='X';
        dfs(x,y+1);
        g[x][y]='.';
        dfs(x,y+1);
    }
}
  
  
int main()
{
    for(int i=0;i<4;i++)
        for(int j=0;j<4;j++)
            cin>>g[i][j];
    dfs(0,0);
    for(int i=0;i<4;i++)
    {
        for(int j=0;j<4;j++)
        {
            if(g[i][j]>='0'&&g[i][j]<='9')cout<<g[i][j];
            else if(cnt[i][j]==d)cout<<'X';
            else if(cnt[i][j]==0)cout<<'O';
            else cout<<'.';
        }
        cout<<endl;
    }
    return 0;
}

python做法

import sys
input = lambda: sys.stdin.readline().rstrip()
ii = lambda: int(input())
lii = lambda: list(map(int, input().split()))
 
dir8 = [(1, 0), (1, 1), (0, 1), (-1, 1), (-1, 0), (-1, -1), (0, -1), (1, -1)]
 
grid = []
for _ in range(4):
    grid.append(list(input()))
     
res = [[0] * 4 for _ in range(4)]
calc = 0
     
def check():
    for r, row in enumerate(grid):
        for c, ch in enumerate(row):
            if ch.isdigit():
                cnt = 0
                for dr, dc in dir8:
                    r2, c2 = r+dr, c+dc
                    if 0 <= r2 < 4 and 0 <= c2 < 4 and grid[r2][c2] == 'X':
                        cnt += 1
                if cnt != int(ch):
                    return False
    return True
     
def dfs(i, j):
    global calc
    if i == 4:
        if check():
            calc += 1
            for r in range(4):
                for c in range(4):
                    res[r][c] += grid[r][c] == 'X'
        return
    if j == 4:
        return dfs(i+1, 0)
    if grid[i][j].isdigit():
        return dfs(i, j+1)
    grid[i][j] = 'X'
    dfs(i, j+1)
    grid[i][j] = '.'
    dfs(i, j+1)
     
dfs(0, 0)
ans = [['.'] * 4 for _ in range(4)]
for r in range(4):
    for c in range(4):
        if grid[r][c].isdigit():
            ans[r][c] = grid[r][c]
        else:
            if res[r][c] == 0:
                ans[r][c] = 'O'
            elif res[r][c] == calc:
                ans[r][c] = 'X'
 
for row in ans:
    print(''.join(row))

5.石子合并

经典的一个区间dp的石子合并问题

c++代码

#include <iostream>
#include <algorithm>
 
using namespace std;
 
const int N = 315;
int s[N];
int dp[N][N], n;
 
int main() {
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> s[i], s[i] += s[i - 1];
 
    for (int len = 2; len <= n; len++) {
        for (int i = 1; i + len - 1 <= n; i++) {
            int j = i + len - 1;
            dp[i][j] = 1e8;
            for (int k = i; k < j; k++) {
                dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j] + s[j] - s[i - 1]);
            }
        }
    }
    printf("%d\n", dp[1][n]);
 
    return 0;
}

python做法

n = int(input())
f = [[0]*(n+2) for _ in range(n+2)]
a = [0] + list(map(int,input().split()))
s = [0]*(n+2)
for i in range(1,n+1):
    s[i] = s[i-1]+a[i]
for le in range(2,n+1):
    i = 1
    while i+le-1<=n:
        l,r=i,i+le-1
        f[l][r] = 1000000000
        for k in range(l,r):
            f[l][r] = min(f[l][r],f[l][k]+f[k+1][r]+s[r]-s[l-1])
        i += 1
print(f[1][n])