十一模拟总结

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在十一国庆期间,我们给祖国过生日,祖国妈妈可高兴了,让gg特意为我们准备了一场模拟!

 

其实这一次的模拟不算太毒瘤,部分分还是可以拿到的,不过考完调正解的时候调到崩溃……

 

T1 matrix

  这道题起手就是30。

  然后我憋了一会儿:这题有一个特别的地方,就是所有询问都在修改之后,那也就应当把所有修改做完,然后快速的询问。

  于是想到了一个类似扫描线的做法,把修改排序,然后维护一棵线段树,区间修改,单点查询,复杂度是O(nqlogn),只能60,结果考试的时候我还算成了O(qlogn),以为能AC……

  正解其实比扫描线简单多了:二维差分!这就是为什么n, m <= 2000了:一是能O(n2),二是能开的下二维数组。于是对于一个矩形的修改(xa, ya)到(xb, yb),我们模仿一维差分,dif[xa][ya]++; dif[xa][yb + 1]--; dif[xb + 1][ya]--; dif[xb + 1][yb + 1]++ 即可(dif是差分数组)。

  最后跑一遍二维前缀和,得到了修改后的矩阵,于是询问的时候就是O(1)的二维前缀和了。

60分代码(线段树区间修改可以改成差分,达到O(nq),然鹅还是60)

技术分享图片
  1 #include<cstdio>
  2 #include<iostream>
  3 #include<cmath>
  4 #include<algorithm>
  5 #include<cstring>
  6 #include<cstdlib>
  7 #include<cctype>
  8 #include<vector>
  9 #include<stack>
 10 #include<queue>
 11 using namespace std;
 12 #define enter puts("") 
 13 #define space putchar(‘ ‘)
 14 #define Mem(a, x) memset(a, x, sizeof(a))
 15 #define rg register
 16 typedef long long ll;
 17 typedef double db;
 18 const int INF = 0x3f3f3f3f;
 19 const db eps = 1e-8;
 20 const int maxn = 2e3 + 5;
 21 inline ll read()
 22 {
 23     ll ans = 0;
 24     char ch = getchar(), last =  ;
 25     while(!isdigit(ch)) {last = ch; ch = getchar();}
 26     while(isdigit(ch)) {ans = ans * 10 + ch - 0; ch = getchar();}
 27     if(last == -) ans = -ans;
 28     return ans;
 29 }
 30 inline void write(ll x)
 31 {
 32     if(x < 0) x = -x, putchar(-);
 33     if(x >= 10) write(x / 10);
 34     putchar(x % 10 + 0);
 35 }
 36 
 37 int n, m, p, q;
 38 
 39 int a[maxn][maxn];
 40 ll Sum[maxn][maxn];
 41 
 42 int l[maxn << 2], r[maxn << 2], lzy[maxn << 2], sum[maxn << 2];
 43 void build(const int& L, const int& R, const int& now)
 44 {
 45     l[now] = L; r[now] = R;
 46     if(L == R) return;
 47     int mid = (L + R) >> 1;
 48     build(L, mid, now << 1);
 49     build(mid + 1, R, now << 1 | 1);
 50 }
 51 void pushdown(const int& now)
 52 {
 53     if(lzy[now])
 54     {
 55         sum[now << 1] += (r[now << 1] - l[now << 1] + 1) * lzy[now];
 56         sum[now << 1 | 1] += (r[now << 1 | 1] - l[now << 1 | 1] + 1) * lzy[now];
 57         lzy[now << 1] += lzy[now];
 58         lzy[now << 1 | 1] += lzy[now];
 59         lzy[now] = 0;
 60     }
 61 }
 62 void update(const int& L, const int& R, const int& now, const int& flg)
 63 {
 64     if(L == l[now] && R == r[now])
 65     {
 66         sum[now] += (R - L + 1) * flg;
 67         lzy[now] += flg; return;
 68     }
 69     pushdown(now);
 70     int mid = (l[now] + r[now]) >> 1;
 71     if(R <= mid) update(L, R, now << 1, flg);
 72     else if(L > mid) update(L, R, now << 1 | 1, flg);
 73     else update(L, mid, now << 1, flg), update(mid + 1, R, now << 1 | 1, flg);
 74     sum[now] = sum[now << 1] + sum[now << 1 | 1];
 75 }
 76 int query(const int& idx, const int& now)
 77 {
 78     if(sum[now] == 0) return 0;
 79     if(l[now] == r[now]) return sum[now];
 80     pushdown(now);
 81     int mid = (l[now] + r[now]) >> 1;
 82     if(idx <= mid) return query(idx, now << 1);
 83     else return query(idx, now << 1 | 1);
 84 }
 85 
 86 struct Node1
 87 {
 88     int L, R, flg;
 89 };
 90 vector<Node1> v1[maxn];
 91 
 92 int main()
 93 {
 94     freopen("matrix.in", "r", stdin);
 95     freopen("matrix.out", "w", stdout);
 96     n = read(); m = read(); p = read(); q = read();
 97     build(1, n, 1);
 98     for(rg int i = 1; i <= p; ++i)
 99     {
100         int xa = read(), ya = read(), xb = read(), yb = read();
101         v1[ya].push_back((Node1){xa, xb, 1}); v1[yb + 1].push_back((Node1){xa, xb, -1});
102     }
103     for(rg int i = 1; i <= m; ++i)
104     {
105         for(rg int j = 0; j < (int)v1[i].size(); ++j)
106             update(v1[i][j].L, v1[i][j].R, 1, v1[i][j].flg);
107         for(rg int j = 1; j <= n; ++j) a[j][i] = query(j, 1);
108     }
109     for(rg int i = 1; i <= n; ++i)
110         for(rg int j = 1; j <= m; ++j) 
111             Sum[i][j] = Sum[i][j - 1] + Sum[i - 1][j] - Sum[i - 1][j - 1] + a[i][j];
112     for(rg int i = 1; i <= q; ++i)
113     {
114         int xa = read(), ya = read(), xb = read(), yb = read();
115         write(Sum[xb][yb] - Sum[xb][ya - 1] - Sum[xa - 1][yb] + Sum[xa - 1][ya - 1]);
116         enter;    
117     }
118     return 0;
119 }
View Code

100分代码

技术分享图片
 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cmath>
 4 #include<algorithm>
 5 #include<cstring>
 6 #include<cstdlib>
 7 #include<cctype>
 8 #include<vector>
 9 #include<stack>
10 #include<queue>
11 using namespace std;
12 #define enter puts("") 
13 #define space putchar(‘ ‘)
14 #define Mem(a, x) memset(a, x, sizeof(a))
15 #define rg register
16 typedef long long ll;
17 typedef double db;
18 const int INF = 0x3f3f3f3f;
19 const db eps = 1e-8;
20 const int maxn = 2e3 + 5;
21 inline ll read()
22 {
23     ll ans = 0;
24     char ch = getchar(), last =  ;
25     while(!isdigit(ch)) {last = ch; ch = getchar();}
26     while(isdigit(ch)) {ans = ans * 10 + ch - 0; ch = getchar();}
27     if(last == -) ans = -ans;
28     return ans;
29 }
30 inline void write(ll x)
31 {
32     if(x < 0) x = -x, putchar(-);
33     if(x >= 10) write(x / 10);
34     putchar(x % 10 + 0);
35 }
36 
37 int n, m, p, q;
38 
39 int a[maxn][maxn], dif[maxn][maxn];
40 ll sum[maxn][maxn];
41 
42 int main()
43 {
44     freopen("matrix.in", "r", stdin);
45     freopen("matrix.out", "w", stdout);
46     n = read(); m = read(); p = read(); q = read();
47     for(rg int i = 1; i <= p; ++i)
48     {
49         int xa = read(), ya = read(), xb = read(), yb = read();
50         dif[xa][ya]++; dif[xa][yb + 1]--; dif[xb + 1][ya]--; dif[xb + 1][yb + 1]++;
51     }
52     
53     for(rg int i = 1; i <= n; ++i)
54         for(rg int j = 1; j <= m; ++j) 
55             a[i][j] = a[i][j - 1] + a[i - 1][j] - a[i - 1][j - 1] + dif[i][j];
56     for(rg int i = 1; i <= n; ++i)
57         for(rg int j = 1; j <= m; ++j) 
58             sum[i][j] = sum[i][j - 1] + sum[i - 1][j] - sum[i - 1][j - 1] + a[i][j];    
59     for(rg int i = 1; i <= q; ++i)
60     {
61         int xa = read(), ya = read(), xb = read(), yb = read();
62         write(sum[xb][yb] - sum[xb][ya - 1] - sum[xa - 1][yb] + sum[xa - 1][ya - 1]);
63         enter;    
64     }
65     return 0;
66 }
View Code

 

T2 card

   O(n3):送的,令dp[i][j]表示第 i 个人选编号为 j 时的方案数,则dp[i][j] = sum(dp[i - 1][h]) (h : 1 ~ m, h + j != k)。然后答案就是sum(dp[n][i]) (i : 1 ~ m)。

  O(n2):维护一个sum[m],表示sum(dp[i]) (i : 1 ~ m),然后就可以省去 h 的一层循环。

  O(n) : 发现,sum[i][m]只跟sum[i - 1][m] 和sum[i - 1][k  - 1]有关,线性dp即可。

  O(logn):矩阵快速幂。然而不会把二维的状态降成一维的,gg。

模拟的时候,推错了O(n)做法,忽视了k > m的情况,得了50.

放一个考场代码吧

技术分享图片
 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cmath>
 4 #include<algorithm>
 5 #include<cstring>
 6 #include<cstdlib>
 7 #include<cctype>
 8 #include<vector>
 9 #include<stack>
10 #include<queue>
11 using namespace std;
12 #define enter puts("") 
13 #define space putchar(‘ ‘)
14 #define Mem(a, x) memset(a, x, sizeof(a))
15 #define rg register
16 typedef long long ll;
17 typedef double db;
18 const int INF = 0x3f3f3f3f;
19 const db eps = 1e-8;
20 const ll mod = 1e9 + 7;
21 inline ll read()
22 {
23     ll ans = 0;
24     char ch = getchar(), last =  ;
25     while(!isdigit(ch)) {last = ch; ch = getchar();}
26     while(isdigit(ch)) {ans = ans * 10 + ch - 0; ch = getchar();}
27     if(last == -) ans = -ans;
28     return ans;
29 }
30 inline void write(ll x)
31 {
32     if(x < 0) x = -x, putchar(-);
33     if(x >= 10) write(x / 10);
34     putchar(x % 10 + 0);
35 }
36 
37 int m, n, k;
38 
39 ll sum, sumk_1;
40 
41 int main()
42 {
43     freopen("card.in", "r", stdin);
44     freopen("card.out", "w", stdout);
45     m = read(); n = read(); k = read();
46     sum = m; sumk_1 = k - 1;
47     for(rg int i = 2; i <= n; ++i)
48     {
49         ll tp = sum;
50         sum = (m * tp % mod - sumk_1 + mod) % mod;
51         sumk_1 = ((k - 1) * tp % mod - sumk_1 + mod) % mod;
52     }
53     write(sum); enter;
54     return 0;
55 }
View Code

 

T3 station

  最暴力的做法是O(n3),然而什么分都得不到。

  对于上述O(n3),枚举车站的时候其实不用再O(n)求一遍,可以用前缀和维护,达到O(n2)。

  在枚举车站的时候,能得到一个很重要的规律,就是当车站向右移动,车站左侧的人的距离增大,右侧的减少,对答案的贡献的变化量就是车站左边人数 - 右边人数。看出总距离是先减少后增大的,因此找最值即可。

  于是80分就有了:用线段树或树状数组维护前缀和,每一次询问相当于单点修改,然后二分查找一个最小的x使得sum(x) >= 1 / 2 * sum(n),复杂度O(qlog2n)。

  100分就是直接有线段树维护,然后再线段树上查询极值点:判断左子区间的sum是否大于等于当前的值,是的话就到左子区间找,否则到右子区间找,直到L == R,返回L即可。

  然后我线段树就gg了,怎么也该不对。

  于是按题解的树状数组倍增写了一发。

模拟时20分的

技术分享图片
 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cmath>
 4 #include<algorithm>
 5 #include<cstring>
 6 #include<cstdlib>
 7 #include<cctype>
 8 #include<vector>
 9 #include<stack>
10 #include<queue>
11 using namespace std;
12 #define enter puts("") 
13 #define space putchar(‘ ‘)
14 #define Mem(a, x) memset(a, x, sizeof(a))
15 #define rg register
16 typedef long long ll;
17 typedef double db;
18 const int INF = 0x3f3f3f3f;
19 const db eps = 1e-8;
20 const int maxn = 2e5 + 5;
21 const ll mod = 998244353;
22 const ll CONST = 19260817;
23 inline ll read()
24 {
25     ll ans = 0;
26     char ch = getchar(), last =  ;
27     while(!isdigit(ch)) {last = ch; ch = getchar();}
28     while(isdigit(ch)) {ans = ans * 10 + ch - 0; ch = getchar();}
29     if(last == -) ans = -ans;
30     return ans;
31 }
32 inline void write(ll x)
33 {
34     if(x < 0) x = -x, putchar(-);
35     if(x >= 10) write(x / 10);
36     putchar(x % 10 + 0);
37 }
38 
39 int n, q;
40 ll a[maxn], sum[maxn], s_mul[maxn];
41 
42 int solve()
43 {
44     ll Min = (ll)INF * (ll)INF;
45     int pos;
46     for(rg int i = 1; i <= n; ++i)
47     {
48         ll tp = ((sum[i] * i) << 1) - (s_mul[i] << 1) + s_mul[n] - sum[n] * i;
49         if(tp < Min) Min = tp, pos = i;
50     }
51     return pos;
52 }
53 
54 ll ans = 0, bas = 1;
55 
56 int main()
57 {
58     freopen("station.in", "r", stdin);
59     freopen("station.out", "w", stdout);
60     n = read(); q = read();
61     for(rg int i = 1; i <= n; ++i) a[i] = read();
62     for(rg int i = 1; i <= q; ++i)
63     {
64         int x = read(); ll b = read();
65         a[x] += b;
66         for(rg int j = 1; j <= n; ++j) sum[j] = sum[j - 1] + a[j], s_mul[j] = s_mul[j - 1] + a[j] * j;
67         (bas *= CONST) %= mod;
68         (ans += bas * solve()) %= mod; 
69     }
70     write(ans); enter;    
71     return 0;
72 }
View Code

100分的

技术分享图片
 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cmath>
 4 #include<algorithm>
 5 #include<cstring>
 6 #include<cstdlib>
 7 #include<cctype>
 8 #include<vector>
 9 #include<stack>
10 #include<queue>
11 using namespace std;
12 #define enter puts("") 
13 #define space putchar(‘ ‘)
14 #define Mem(a, x) memset(a, x, sizeof(a))
15 #define rg register
16 typedef long long ll;
17 typedef double db;
18 const int INF = 0x3f3f3f3f;
19 const db eps = 1e-8;
20 const ll mod = 998244353;
21 const ll CONST = 19260817;
22 const int maxn = (1 << 20) + 5;
23 inline ll read()
24 {
25     ll ans = 0;
26     char ch = getchar(), last =  ;
27     while(!isdigit(ch)) {last = ch; ch = getchar();}
28     while(isdigit(ch)) {ans = ans * 10 + ch - 0; ch = getchar();}
29     if(last == -) ans = -ans;
30     return ans;
31 }
32 inline void write(ll x)
33 {
34     if(x < 0) x = -x, putchar(-);
35     if(x >= 10) write(x / 10);
36     putchar(x % 10 + 0);
37 }
38 
39 int n, p;
40 
41 ll c[maxn], sum = 0;
42 int lowbit(int x)
43 {
44     return x & -x;
45 }
46 void add(int pos, ll d)
47 {
48     for(; pos < maxn; pos += lowbit(pos)) c[pos] += d; 
49     sum += d;
50 }
51 int query(ll x)
52 {
53     int pos = 0; ll res = 0;
54     for(int i = (1 << 17); i; i >>= 1) if(res + c[pos + i] <= x) pos += i, res += c[pos];
55     return pos + 1;
56 }
57 
58 ll ans = 0, bas = 1;
59 
60 int main()
61 {
62     freopen("station.in", "r", stdin);
63     freopen("station.out", "w", stdout);
64     n = read(); p = read();
65     for(int i = 1; i <= n; ++i) {ll x = read(); add(i, x);}
66     for(int i = 1; i <= p; ++i) 
67     {
68         int x = read(); ll d = read();
69         add(x, d); 
70         bas = bas * CONST % mod;
71         ans = (ans + bas * query((sum - 1) >> 1)) % mod;
72     }
73     write(ans); enter;
74     return 0;
75 }
View Code

 

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