2019.10.03解题报告

Posted 2022-12-19 loceaner

总体来说能打的暴力都打了
期望\(100 + 40 + 30 = 170\)
实际\(100 + 40 + 10 = 180\)
数据良心（其实是数据太水惹）

T1

第一眼觉得就是要找规律，然后直接找找不出来，所以用暴力搜一下

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream> 
using namespace std;

const int N = 2011;

int vis[N][N];

struct node 
    int x, y, num;
;

const int dx[8] = -2,-2,-1,1,2,2,1,-1;
const int dy[8] = -1,1,2,2,1,-1,-2,-2;

queue <node> q;

int sum = 0;

int main() 
    for(int T = 0; T <= 100; T++) 
        int n = T;
        sum = 0;
        memset(vis, 0, sizeof(vis));
        while(!q.empty()) q.pop();
        node st;
        st.x = 1000;
        st.y = 1000;
        vis[st.x][st.y] = 1;
        st.num = 0;
        q.push(st);
        while(!q.empty()) 
            node top = q.front();
            q.pop();
            sum++;
            for(int i = 0; i < 8; i++) 
                node tmp;
                tmp.x = top.x + dx[i]; 
                tmp.y = top.y + dy[i];
                tmp.num = top.num + 1;
                if(!vis[tmp.x][tmp.y] && tmp.x >= 0 && tmp.y >= 0 && tmp.x <= 2000 && tmp.y <= 2000 &&tmp.num <= n) 
                    vis[tmp.x][tmp.y] = tmp.num;
                    q.push(tmp);
                
            
        
        cout  << sum << '\n';
    

找不到什么规律，于是求一下差值，还是没什么规律，用差值再求一下差值，发现前五项的差值是特别的，所以直接特别输出，其他的发现差值恒定为\(28\)，规律就找到啦

\(if(n > 5)\ n \ \ -= 4\)

\(a\)数组表示原数列
\(a[1] = 325\)
\(a[2] = 473\)
\(a[3] = 649\)
\(a[4] = 853\)
\(a[5] = 1085\)

\(b\)数组表示差
\(b[1] = 148\)
\(b[2] = 176\)
\(b[3] = 204\)
\(b[4] = 232\)
\(.................\)
\(b[n] = 120 + n * 28\)

显然
\(b[2] = b[1] + 28\)
\(b[3] = b[2] + 28\)
\(....................................\)
\(b[n] = b[n - 1] + 28\)

有
\(a[2] - a[1] = b[1]\)
\(a[3] - a[2] = b[2]\)
\(a[4] - a[3] = b[3]\)
\(a[5] - a[4] = b[4]\)
\(.............................\)
\(a[n] - a[n - 1] = b[n - 1]\)

加起来
\(a[2] - a[1] + a[3] - a[2] + a[4] - a[3] + .......... a[n] - a[n - 1] = b[1] + b[2] + b[3] + ......... b[n - 1]\)

去掉多余项，得
\(a[n] - a[1] = \sum_i = 1^n - 1 b[i]\)
设\(x = \sum_i = 1^n - 1 b[i]\)

所以\(a[n] = x + a[1]\)
x是个等差数列，然后等差数列求和公式为\(\frac首项 + 尾项2 *数量\)

这里就是\(\fracb[1] + b[n - 1]2 *(n - 1)\)

所以就是\(\fracb[1] + b[n - 1]2 *(n - 1) + a[1]\)

就等于\(\frac120 + 28 + 120 + 28 * （n - 1）2 *(n - 1) + 325\)

然后化简一下，可能爆\(long\ long\)，保险起见用\(unsigned\ long\ long\)，然后就做完了

//知识点:宽搜打表找规律 
/*
By:Loceaner
找到神奇规律之后发现是等差数列…… 
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#define ull unsigned long long
using namespace std;

int T;
ull n;
ull a[5] = 1, 9, 41, 109, 205;

int main() 
    freopen("horse.in", "r", stdin);
    freopen("horse.out", "w", stdout); 
    scanf("%d", &T);
    while(T--) 
        cin >> n;
        if(n < 5) 
            cout << a[n] << "\n";
            continue;
        
        else 
            n -= 4;
            ull ans = 120 * (n - 1) + 14 * n * (n - 1) + 325;
//          ull ans = (240 + 28 * n) / 2 * (n - 1) + 325;
            cout << ans << '\n';
        
    
    return 0;

T2

很显然，第一问的答案就是 \(2^n\)。
第二问的话。。到现在也不会
然后dsr聚聚给我讲了半年，我也不会，上聚聚的代码

#include <bits/stdc++.h>
using namespace std;
const int _=1e5+7;
int n,len,k,vis[_],ans[_];
int main() 
    freopen("sensor.in","r",stdin);
    freopen("sensor.out","w",stdout);
    cin>>k;n=1<<k;
    vis[0]=1;
    for(int i=n-k+1;i<=n;++i) ans[i]=1;
    for(int i=0;i<k;++i) vis[n-(1<<i)]=1;
    for(int i=k+1,jt=0;i<=n-k;++i) 
        jt=(jt<<1)&(n-1);
        if(vis[jt]) ans[i]=1,jt|=1;
        vis[jt]=1;
    
    printf("%d ",n);
    for(int i=1;i<=n;++i) printf("%d",ans[i]);
    return 0;

T3

第一眼觉得是线段树，因为一列的话挺好操作，可是有十五列，开十五个线段树？？

懵逼了

再加上没有时间写了，于是果断写了暴力，拿了40

下面是40分暴力和修改过的满分代码

//暴力啦啦啦
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

inline int read() 
    char c = getchar();
    int x = 0, f = 1;
    for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
    return x * f;


int r, c, q;
char a[50011][20];
char b[50011][20];

int main() 
    freopen("holiday.in", "r", stdin);
    freopen("holiday.out", "w", stdout);
    r = read(), c = read(), q = read();
    for(int i = 1; i <= r; i++) 
        scanf("%s",a[i]+1);
    
    for(int i = 1; i <= r; i++) 
        for(int j = 1; j <= c; j++) 
            b[i][j] = '0';
        
    
    while(q--) 
        int cnt = 0;
        int u1 = read(), v1 = read(), u2 = read(), v2 = read();
        char col;
        cin >> col;
        for(int i = u1; i <= v1; i++) 
            for(int j = u2; j <= v2; j++) 
                b[i][j] = col;
            
        
        for(int i = 1; i <= r; i++) 
            for(int j = 1; j <= c; j++) 
                if(a[i][j] == b[i][j]) cnt++;
            
        
        cout << cnt << '\n';
    
    return 0;

修改后

//知识点:多个线段树 
/*
By:Loceaner
第一眼线段树，第二眼不会打
第三眼打暴力，最后多了10分 
数据良心 
*/
#include <cstdio>
#include <cstring> 
#include <iostream>
#include <algorithm>
using namespace std;

const int N = 5e4 + 11;

inline int read() 
    char c = getchar();
    int x = 0, f = 1;
    for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
    return x * f;


char a[N][20];
int r, c, q;

#define lson rt << 1
#define rson rt << 1 | 1

struct seg_tree
    struct node 
        int siz, tot, ans, tag;
     t[N << 2];
    
    void build(int l, int r, int rt, int k) 
        t[rt].tag = -1;
        t[rt].siz = r - l + 1;
        if(l == r) 
            t[rt].tot = (a[l][k] == '1');
            t[rt].ans = (a[l][k] == '0');
            return;
        
        int mid = (l + r) >> 1;
        build(l, mid, lson, k);
        build(mid + 1, r, rson, k);
        t[rt].tot = t[lson].tot + t[rson].tot;
        t[rt].ans = t[lson].ans + t[rson].ans;
    
    
    void tag(int rt, int ad) 
        if(ad == 1) t[rt].ans = t[rt].tot;
        else t[rt].ans = t[rt].siz - t[rt].tot;
        t[rt].tag = ad;
    
    
    void pushdown(int rt) 
        if(t[rt].tag != -1) 
            tag(lson, t[rt].tag);
            tag(rson, t[rt].tag);
            t[rt].tag = -1;
        
    
    
    void update(int L, int R, int k, int l, int r, int rt) 
        if(L <= l && r <= R) 
            tag(rt, k);
            return;
        
        pushdown(rt);
        int mid = (l + r) >> 1;
        if(L <= mid) update(L, R, k, l, mid, lson);
        if(R > mid) update(L, R, k, mid + 1, r, rson);
        t[rt].ans = t[lson].ans + t[rson].ans;
    
 T[16];

int main() 
    freopen("holiday.in", "r", stdin);
    freopen("holiday.out", "w", stdout);
    r = read(), c = read(), q = read();
    for(int i = 1; i <= r; i++) scanf("%s", a[i] + 1);
    for(int i = 1; i <= c; i++) T[i].build(1, r, 1, i);
    while(q--) 
        int L = read(), R = read(), u2 = read(), v2 = read(), col = read();
        for(int i = u2; i <= v2; i++) T[i].update(L, R, col, 1, r, 1);
        int ans = 0;
        for(int i = 1; i <= c; i++) ans += T[i].t[1].ans;   
        cout << ans << '\n';
    
    return 0;