NOJ AC25 记录

OJ刷题总结

NOJ1009 2的n次方

模拟小学时候的竖乘，注意code中extra变量的位置

#include <iostream>

using namespace std;

const int N = 1000;
int n, a[N];

int main()
{
    int k = 1;
    cin >> n;
    a[1] = 1;
    for(int i = 1; i <= n; i++ )
    {
        int extra = 0;
        for(int j = 1; j <= k; j++ )
        {
            a[j] = (a[j] << 1) + extra;
            extra = a[j] / 10;
            a[j] %= 10;
            if(extra != 0 && j == k) k++;
        }
    }

    for(int i = k; i >= 1; i-- )
        printf("%d", a[i]);
    puts("");

    return 0;
}

5.NOJ1011 大数加法

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 1010;
string s1, s2;
int a[N], b[N];
int ans[N];

void switch_toint(string s, int u[])
{
    for(int i = 0; i < s.length(); i++ )
        u[i] = s[s.length() - i - 1] - '0';
}

int main()
{
    cin >> s1 >> s2;
    switch_toint(s1, a);
    switch_toint(s2, b);
    
    int edge = max(s1.length(), s2.length());
    
    for(int i = 0; i < edge; i++ )
    {
        ans[i] += a[i] + b[i];
        ans[i + 1] = ans[i] / 10;
        ans[i] %= 10;
    }
    
    if(ans[edge])
        printf("%d", ans[edge]);
    for(int i = edge - 1; i >= 0; i-- )
        printf("%d", ans[i]);
    puts("");
    
    return 0;
}

NOJ1012 进制转换

特别基础的题，就是因为忘记特判n == 0 然后wa卡了5分钟，要认真审题要认真审题要认真审题

#include <iostream>
#include <cstdio>

using namespace std;

char a[17] = "0123456789ABCDEF";

void dfs(int n, int r)
{
    if(!n) return;
    dfs(n / r, r);
    cout << a[n % r];
}

int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        int n, r;
        scanf("%d%d", &n, &r);
        if(n < 0) cout <<'-';
        if(!n){cout << '0' << endl; continue;}
        dfs(abs(n), r);
        puts("");
    }
    
    return 0;
}

NOJ1142 最大连续和

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

const int N = 1e6 + 10;
LL a[N], s[N];

int main()
{
    int t;
    scanf("%d", &t);
    
    while(t--)
    {
        int m;
        scanf("%d", &m);
        
        memset(s, 0, sizeof s);
        for(int i = 1; i <= m; i++ )
        {
            scanf("%lld", a + i);
            s[i] = s[i - 1] + a[i];
        }
        
        LL res = -1e10;
        for(int i = 0; i <= m; i++ )
            for(int j = i + 1; j <= m; j++ )
                res = max(res, s[j] - s[i]);
        
        printf("%lld
", res);
    }
    
    return 0;
}

NOJ1094 蛇形填数

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 70;
int a[N][N];

int main()
{
    int n, k = 1;
    cin >> n;
    
    int i = 0, j = n - 1;
    a[i][j] = 1;
    while(k < n * n)
    {
        while(a[i + 1][j] == 0 && i <= n - 2) a[++i][j] = ++k;
        while(a[i][j - 1] == 0 && j >= 1) a[i][--j] = ++k;
        while(a[i - 1][j] == 0 && i >= 1) a[--i][j] = ++k;
        while(a[i][j + 1] == 0 && j <= n - 2) a[i][++j] = ++k;
    }
    
    for(int i = 0; i < n; i++ )
    {
        for(int j = 0; j < n; j++ )
            printf("%5d", a[i][j]);
        cout << endl;
    }
    return 0;
    
}

HDU1009 FatMouse Trade

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 1010;
struct room
{
    int jb, cf;
    double rate;
    
    bool operator< (const room & t)const
    {
        return rate > t.rate;
    }
}g[N];
int m, n;

int main()
{
    while(scanf("%d%d", &m, &n), m + 1)
    {
        for(int i = 0; i < n; i++ )
        {
            scanf("%d%d", &g[i].jb, &g[i].cf);
            g[i].rate = g[i].jb * 1.0 / g[i].cf;
        }
        
        sort(g, g + n);
        
        
        double sum = 0;
        for(int i = 0; i < n; i++ )
        {
            if(m <= 0) break;
            if(m >= g[i].cf)
            {
                sum += g[i].jb;
                m -= g[i].cf;
                continue;
            }
            if(m < g[i].cf && m > 0)
            {
                sum += g[i].rate * m;
                m = 0;
            }
        }
        
        printf("%.3lf
", sum);
    }
    return 0;
}

• 一些字符串处理的问题

HDU2072 单词数

计算不重复的单词出现个数，很容易想到set容器，然后用stringstream处理字符流输出set的尺寸即为答案

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <set>
#include <sstream>

using namespace std;

int main()
{
    string s;
    
    while(getline(cin, s), s != "#")
    {
        set<string> words;
        stringstream ssin(s);
        string u;
        
        while(ssin >> u)
            words.insert(u);
            
        cout << words.size() << endl;
    }
    
    return 0;
}

HDU2081 手机短号

通过stringstream来实现字符串与整数间的转换

#include <cstring>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <sstream>

using namespace std;

typedef long long LL;

int main()
{
    int t;
    cin >> t;

    while(t--)
    {
        string s;
        LL res;
        cin >> s;

        stringstream ssin(s);
        ssin >> res;
        printf("%lld
", 600000 + res % 100000);
    }
    return 0;
}

NOJ1023 字符串排序

题目表述有歧义，，，

比较基础的一题，重载一个运算符就可以ac

#include <bits/stdc++.h>

#define x first
#define y second

using namespace std;

typedef pair<string, int> PSI;

const int N = 1e5 + 10;
PSI str[N];

int cnt(string s)
{
    int res = 0;
    for(auto x : s)
        if(x == 'A') res++;
    
    return res;
}

bool cmp(const PSI & s1, const PSI & s2)
{
    if(cnt(s1.x) < cnt(s2.x)) return true;
    else if(cnt(s1.x) > cnt(s2.x)) return false;
    else
    {
        if(s1.y < s2. y) return true;
        else return false;
    }
}

int main()
{
    int res = 1;
    while(cin >> str[res].x)
    {
        str[res].y = res;
        res++;
    }
    
    sort(str + 1, str + res, cmp);
    
    for(int i = 1; i < res; i++ )
        cout << str[i].x << endl;
        
    return 0;
    
}

HDU2602 01背包问题

简单的dp，

DP核心就是状态表示 + 状态计算(即寻找状态转移方程)

数组的某个元素表示的是某个选法的所有集合，存放的值即元素的属性是我们所需要的，一般分为三类(max, min, cnt)一般从最后一步来思考状态转移方程。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>

using namespace std;

const int N = 1010;

int main()
{
    int t;
    cin >> t;
    while(t--)
    {
        int n, m;
        int v[N], w[N];
        int f[N][N];

        memset(f, 0, sizeof f);

        cin >> n >> m;
        for(int i = 1; i <= n; i++ )
            cin >> w[i];
        for(int i = 1; i <= n; i++ )
            cin >> v[i];
        
        for(int i = 1; i <= n; i++ )
            for(int j = 0; j <= m; j++ )
            {
                f[i][j] = f[i - 1][j];
                if(j >= v[i]) f[i][j] = max(f[i][j], f[i - 1][j - v[i]] + w[i]);
            }
        cout << f[n][m] << endl;
    }
    return 0;
}

HDU1003 Max sum

第一遍写的dp，内存过大，超时

状态转移方程想的太过复杂，写出来就像是暴力算法，不可取。

正确的状态转移方程sum[i - 1] >= 0 ? sum[i] = sum[i - 1] + a[i] : sum[i] = 0;

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 1010;
int a[N], f[N][N];

int main()
{
    int t;
    scanf("%d", &t);
    for(int u = 1; u <= t; u++ )
    {
        int m, res = 0;
        memset(f, 0, sizeof f);
        
        scanf("%d", &m);
        for(int i = 1; i <= m; i++ ) scanf("%d", a + i);
        
        for(int i = 1; i <= m; i++ )
            for(int j = i; j <= m; j++ )
            {
                f[i][j] = f[i][j - 1] + a[j];
                res = max(res, f[i][j]);
            }
        int flag = 1;
        for(int i = 1; i <= m && flag; i++ )
            for(int j = 1; j <= m; j++ )
                if(f[i][j] == res)
                {
                    printf("Case %d:
", u);
                    printf("%d %d %d

", f[i][j], i, j);
                    flag = 0;
                    break;
                }
    }
    
    return 0;
}

//ACcode
//from csdn
#include<stdio.h>
int main()
{
  int n,m,t,i,j;
  scanf("%d",&t);
  int num=1;
  while(t--)
  {
    int max=-9999;
    int sum=0;
    int begin=1,end=1,x=1;
    scanf("%d",&n);
    for(i=0; i<n; i++)
    {
        scanf("%d",&m);
        sum+=m;//记录最大连续和的值
        if(sum>max)
        {
            max=sum;
            begin=x;//记录开始于第几个数
            end=i+1;//记录结束于第几个数
        }
        if(sum<0)//都是负数的情况
        {
            x=i+2;//注意：这时候是从下一个输入的数开始计算
            sum=0;
        }
    }
    if(num!=1)
    printf("
");//注意输出格式
    printf("Case %d:
",num++);
    printf("%d %d %d
",max,begin,end);
  }
  return 0;
}

NOJ1030 ACM程序设计之马拉松竞赛

被空格卡的难受

因为第一遍代码写的不够周全，被一个刚开始没想到的样例卡了好久。

1.全部领奖 2.全部不领奖

HDU1166 敌兵布阵

wa的代码输入方式和ac的不一样，要注意！

考察树状数组，难度较小

//WAcode
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long LL;

const int N = 50010;
int tr[N];
int n, val;

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

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

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

int main()
{
    int t;
    scanf("%d", &t);
    for(int u = 1; u  <= t; u++ )
    {
        memset(tr, 0, sizeof tr);
        
        scanf("%d", &n);
        for(int i = 1; i <= n; i++ )
        {
            scanf("%d", &val);
            add(i, val);
        }
        
        printf("Case %d:
", u);
        
        string c;
        int j, k;
        
        while(cin >> c >> j >> k && c[0] != 'E')
        {
            if(c[0] == 'Q')
                cout << query(k) - query(j - 1) << endl;
            else if(c[0] == 'A')
                add(j, k);
            else if(c[0] == 'S')
                add(j, -k);
        }
        
    }
    return 0;
}

//ACcode
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long LL;

const int N = 50010;
int tr[N];
int n, val;

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

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

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

int main()
{
    int t;
    scanf("%d", &t);
    for(int u = 1; u  <= t; u++ )
    {
        memset(tr, 0, sizeof tr);
        
        scanf("%d", &n);
        for(int i = 1; i <= n; i++ )
        {
            scanf("%d", &val);
            add(i, val);
        }
        
        printf("Case %d:
", u);
        
        char c[10];
        int j, k;
        scanf("%s", c);
        while(c[0] != 'E')
        {
            if(c[0] == 'Q')
            {
                scanf("%d%d", &j, &k);
                cout << query(k) - query(j - 1) << endl;
            }
            else if(c[0] == 'A')
            {
                scanf("%d%d", &j, &k);
                add(j, k);
            }
            else if(c[0] == 'S')
            {
                scanf("%d%d", &j, &k);
                add(j, -k);
            }
            scanf("%s", c);
        }
        
    }
    return 0;
}