最后一题是Splay...还没有学会。。蒟蒻！！！
A
1 /*************************************************************************
2 > File Name: A.cpp
3 > Author: Stomach_ache
4 > Mail: [email protected]
5 > Created Time: 2014年09月28日 星期日 13时33分32秒
6 > Propose:
7 ************************************************************************/
8
9 #include <cmath>
10 #include <string>
11 #include <cstdio>
12 #include <fstream>
13 #include <cstring>
14 #include <iostream>
15 #include <algorithm>
16 using namespace std;
17 /*Let‘s fight!!!*/
18
19 int n, A[100005];
20
21 int main(void) {
22 ios::sync_with_stdio(false);
23 int t;
24 cin >> t;
25 while (t--) {
26 cin >> n;
27 for (int i = 0; i < n; i++) cin >> A[i];
28 sort(A, A + n);
29 long long res = 0;
30 for (int i = n - 1; i >= 0; i -= 2) res += A[i];
31 cout << res << endl;
32 }
33 return 0;
34 }
B
处理出每个素数在所有数中出现的次数最多的个数。
1 /*************************************************************************
2 > File Name: B.cpp
3 > Author: Stomach_ache
4 > Mail: [email protected]
5 > Created Time: 2014年09月28日 星期日 13时36分46秒
6 > Propose:
7 ************************************************************************/
8 #include <map>
9 #include <cmath>
10 #include <string>
11 #include <cstdio>
12 #include <vector>
13 #include <fstream>
14 #include <cstring>
15 #include <iostream>
16 #include <algorithm>
17 using namespace std;
18 /*Let‘s fight!!!*/
19
20 typedef pair<int, int> pii;
21 const int MAX_N = 100005;
22 const int MAX_M = 1000005;
23 bool vis[MAX_M];
24 int prime[MAX_N], A[MAX_N], cnt;
25
26 void init() {
27 cnt = 0;
28 memset(vis, false, sizeof(vis));
29 for (int i = 2; i < MAX_M; i++) {
30 if (!vis[i]) {
31 prime[++cnt] = i;
32 for (int j = 2*i; j < MAX_M; j += i) vis[j] = true;
33 }
34 }
35 }
36
37 #define rep(i, n) for (int i = (1); i <= (n); i++)
38
39 vector<pii> factor(MAX_M);
40 vector<pii>::iterator it;
41
42 void work(int x, int y) {
43 if (factor[x].second == 0) {
44 factor[x].second = y; }
45 else {
46 factor[x].second = max(y, factor[x].second);
47 }
48 }
49
50 int main(void) {
51 init();
52 ios::sync_with_stdio(false);
53 int t, n;
54 cin >> t;
55 while (t--) {
56 rep (i, MAX_M) factor[i-1].second = 0;
57 cin >> n;
58 rep (i, n) cin >> A[i];
59 rep (i, n) {
60 int x = A[i];
61 if (x == 1) continue;
62 rep (j, cnt) {
63 if (prime[j]*prime[j] > A[i]) break;
64 if (x % prime[j] == 0) {
65 int tmp = 0;
66 while (x % prime[j] == 0) tmp++, x /= prime[j];
67 work(prime[j], tmp);
68 }
69 }
70 if (x > 1) work(x, 1);
71 }
72 int res = 0;
73 for (it = factor.begin(); it != factor.end(); ++it) {
74 res += it->second;
75 }
76 cout << res << endl;
77 }
78
79 return 0;
80 }
C
线段树。
对于第一种修改，就是区间减１
对于第二种修改，就是单点更新
最后，查询每个点２，３，５因子的个数。
1 /*************************************************************************
2 > File Name: C.cpp
3 > Author: Stomach_ache
4 > Mail: [email protected]
5 > Created Time: 2014年09月28日 星期日 14时06分24秒
6 > Propose:
7 ************************************************************************/
8 #include <cmath>
9 #include <string>
10 #include <cstdio>
11 #include <fstream>
12 #include <cstring>
13 #include <iostream>
14 #include <algorithm>
15 using namespace std;
16 /*Let‘s fight!!!*/
17
18 const int MAX_N = 100050;
19 int factor[3][MAX_N], A[MAX_N], id[MAX_N];
20 #define rep(i, n) for (int i = (1); i <= (n); i++)
21 #define lson(x) ((x<<1))
22 #define rson(x) ((x<<1) | 1)
23
24 struct node {
25 int l, r, Min[3];
26 }Tr[MAX_N<<2];
27
28 void build(int rt, int l, int r) {
29 Tr[rt].l = l, Tr[rt].r = r;
30 rep(i, 3) Tr[rt].Min[i-1] = 0;
31 if (l == r) {
32 rep (i, 3) Tr[rt].Min[i-1] = factor[i-1][l];
33 id[l] = rt;
34 return ;
35 }
36 int mid = (l + r) / 2;
37 build(lson(rt), l, mid);
38 build(rson(rt), mid + 1, r);
39 }
40
41 void pushdown(int rt, int p) {
42 if (Tr[rt].Min[p] != 0) {
43 Tr[lson(rt)].Min[p] += Tr[rt].Min[p];
44 Tr[rson(rt)].Min[p] += Tr[rt].Min[p];
45 Tr[rt].Min[p] = 0;
46 }
47 }
48
49 void update1(int rt, int l, int r, int p) {
50 if (Tr[rt].l >= l && Tr[rt].r <= r) {
51 Tr[rt].Min[p]--;
52 return ;
53 }
54 pushdown(rt, p);
55 int mid = Tr[lson(rt)].r;
56 if (l <= mid) update1(lson(rt), l, r, p);
57 if (r > mid) update1(rson(rt), l, r, p);
58 }
59
60 void update2(int rt, int l, int p, int d) {
61 if (Tr[rt].l == Tr[rt].r && Tr[rt].l == l) {
62 Tr[rt].Min[p] = d;
63 return ;
64 }
65 pushdown(rt, p);
66 int mid = Tr[lson(rt)].r;
67 if (l <= mid) update2(lson(rt), l, p, d);
68 else update2(rson(rt), l, p, d);
69 }
70
71 int query(int rt, int l, int p) {
72 if (Tr[rt].l == Tr[rt].r && Tr[rt].l == l) {
73 return max(Tr[rt].Min[p], 0);
74 }
75 pushdown(rt, p);
76 int mid = Tr[lson(rt)].r;
77 if (l <= mid) query(lson(rt), l, p);
78 else query(rson(rt), l, p);
79 }
80
81 void read(int &res) {
82 res = 0;
83 char c = ‘ ‘;
84 while (c < ‘0‘ || c > ‘9‘) c = getchar();
85 while (c >= ‘0‘ && c <= ‘9‘) res = res*10+c-‘0‘, c = getchar();
86 }
87
88 int POW(int a, int b) {
89 int res = 1;
90 while (b) {
91 if (b & 1) res *= a;
92 a *= a;
93 b >>= 1;
94 }
95 return res;
96 }
97
98 int main(void) {
99 int N, M, a[4] = {0, 2, 3, 5};
100 read(N);
101 rep (i, N) {
102 read(A[i]);
103 int x = A[i];
104 factor[0][i] = factor[1][i] = factor[2][i] = 0;
105 rep (j, 3) {
106 int tmp = 0, p = a[j];
107 while (x % p == 0) tmp++, x /= p;
108 factor[j-1][i] = tmp;
109 }
110 A[i] = x;
111 }
112 build(1, 1, N);
113 read(M);
114 int t, l, r, p, d, tmp;
115 while (M--) {
116 read(t);
117 if (t == 1) {
118 read(l), read(r), read(p);
119 update1(1, l, r, (p+1)/2-1);
120 } else {
121 read(l), read(d);
122 rep (i, 3) {
123 tmp = 0, p = a[i];
124 while (d % p == 0) tmp++, d /= p;
125 update2(1, l, i-1, tmp);
126 }
127 A[l] = d;
128 }
129 }
130 rep (i, N) {
131 rep (j, 3) {
132 tmp = query(1, i, j-1);
133 A[i] *= POW(a[j], tmp);
134 }
135 printf("%d%c", A[i], i == N ? ‘\n‘ : ‘ ‘);
136 }
137
138 return 0;
139 }
D
时间： 2024-08-01 10:42:02