题目链接：
http://acm.hdu.edu.cn/listproblem.php?vol=53
A题：
题意：
总共有sum(a[i])篇文章，文章含有i条引用的文章数是ai，求最大的h使得最少有h篇文章含有至少h条引用。
思路：
二分，不过一开始把i和ai的含义读反wa了几发。
代码实现如下：
1 #include <set>
2 #include <map>
3 #include <deque>
4 #include <queue>
5 #include <stack>
6 #include <cmath>
7 #include <ctime>
8 #include <bitset>
9 #include <cstdio>
10 #include <string>
11 #include <vector>
12 #include <cstdlib>
13 #include <cstring>
14 #include <iostream>
15 #include <algorithm>
16 using namespace std;
17
18 typedef long long LL;
19 typedef pair<LL, LL> pLL;
20 typedef pair<LL, int> pLi;
21 typedef pair<int, LL> pil;;
22 typedef pair<int, int> pii;
23 typedef unsigned long long uLL;
24
25 #define lson rt<<1
26 #define rson rt<<1|1
27 #define lowbit(x) x&(-x)
28 #define name2str(name) (#name)
29 #define bug printf("*********\n")
30 #define debug(x) cout<<#x"=["<<x<<"]" <<endl
31 #define FIN freopen("D://code//in.txt","r",stdin)
32 #define IO ios::sync_with_stdio(false),cin.tie(0)
33
34 const double eps = 1e-8;
35 const int mod = 1000000007;
36 const int maxn = 2e5 + 7;
37 const double pi = acos(-1);
38 const int inf = 0x3f3f3f3f;
39 const LL INF = 0x3f3f3f3f3f3f3f3fLL;
40
41 int n;
42 int a[maxn];
43
44 bool check(LL x) {
45 LL sum = 0;
46 for(int i = 0; i <= n; i++) {
47 if(i >= x) sum += a[i];
48 }
49 return sum >= x;
50 }
51
52 int main(){
53 while(~scanf("%d", &n)) {
54 for(int i = 0; i <= n; i++) {
55 scanf("%d", &a[i]);
56 }
57 LL ub = INF, lb = 0, mid, ans = 0;
58 while(ub >= lb) {
59 mid = (ub + lb) >> 1;
60 if(check(mid)) {
61 lb = mid + 1;
62 ans = mid;
63 } else {
64 ub = mid - 1;
65 }
66 }
67 printf("%lld\n", ans);
68 }
69 return 0;
70 }
B题：
题意：
你一共有n个小时，如果你花x个小时写一篇文章，那么这篇文章将会有x*a条引用，问你求h-index。
思路：
看样例大概可以知道每篇文章最多可以引用一次，我们容易发现我们只有每篇文章都只用1小时是最优的，则此时所写出的n篇文章所含有的引用条数分别为a,a+1,a+2……a+n-1。
我们假设答案是x，显然x>=a，则它在数组中的位置为x-a+1（下标从1开始），则数组中大于等于x的数共有n-x+a,由h-index的定义知，n-x+a>=x，咕x<=(n+a)/2，最大值就是(n+a)/2。
代码实现如下：
1 #include <set>
2 #include <map>
3 #include <deque>
4 #include <queue>
5 #include <stack>
6 #include <cmath>
7 #include <ctime>
8 #include <bitset>
9 #include <cstdio>
10 #include <string>
11 #include <vector>
12 #include <cstdlib>
13 #include <cstring>
14 #include <iostream>
15 #include <algorithm>
16 using namespace std;
17
18 typedef long long LL;
19 typedef pair<LL, LL> pLL;
20 typedef pair<LL, int> pLi;
21 typedef pair<int, LL> pil;;
22 typedef pair<int, int> pii;
23 typedef unsigned long long uLL;
24
25 #define lson rt<<1
26 #define rson rt<<1|1
27 #define lowbit(x) x&(-x)
28 #define name2str(name) (#name)
29 #define bug printf("*********\n")
30 #define debug(x) cout<<#x"=["<<x<<"]" <<endl
31 #define FIN freopen("D://code//in.txt","r",stdin)
32 #define IO ios::sync_with_stdio(false),cin.tie(0)
33
34 const double eps = 1e-8;
35 const int mod = 1000000007;
36 const int maxn = 2e5 + 7;
37 const double pi = acos(-1);
38 const int inf = 0x3f3f3f3f;
39 const LL INF = 0x3f3f3f3f3f3f3f3fLL;
40
41 int n, a;
42
43 int main(){
44 while(~scanf("%d%d", &n, &a)) {
45 printf("%d\n", (n + a) / 2);
46 }
47 return 0;
48 }
C题：
题意：
求[Li,Ri]中的h-index。
思路：
对于每次查询，我们只需要二分答案x，然后看第x+1大的数是否大于等于x即可，主席树裸题。
代码实现如下：
1 #include <set>
2 #include <map>
3 #include <deque>
4 #include <queue>
5 #include <stack>
6 #include <cmath>
7 #include <ctime>
8 #include <bitset>
9 #include <cstdio>
10 #include <string>
11 #include <vector>
12 #include <cstdlib>
13 #include <cstring>
14 #include <iostream>
15 #include <algorithm>
16 using namespace std;
17
18 typedef long long LL;
19 typedef pair<LL, LL> pLL;
20 typedef pair<LL, int> pLi;
21 typedef pair<int, LL> pil;;
22 typedef pair<int, int> pii;
23 typedef unsigned long long uLL;
24
25 #define lson rt<<1
26 #define rson rt<<1|1
27 #define lowbit(x) x&(-x)
28 #define name2str(name) (#name)
29 #define bug printf("*********\n")
30 #define debug(x) cout<<#x"=["<<x<<"]" <<endl
31 #define FIN freopen("D://code//in.txt","r",stdin)
32 #define IO ios::sync_with_stdio(false),cin.tie(0)
33
34 const double eps = 1e-8;
35 const int mod = 1000000007;
36 const int maxn = 1e5 + 7;
37 const double pi = acos(-1);
38 const int inf = 0x3f3f3f3f;
39 const LL INF = 0x3f3f3f3f3f3f3f3fLL;
40
41 int n, q, cnt, x, y;
42 int a[maxn], root[maxn];
43
44 struct node {
45 int l, r, sum;
46 }tree[maxn*40];
47
48 void update(int l, int r, int& x, int y, int pos) {
49 tree[++cnt] = tree[y], tree[cnt].sum++, x = cnt;
50 if(l == r) return;
51 int mid = (l + r) >> 1;
52 if(mid >= pos) update(l, mid, tree[x].l, tree[y].l, pos);
53 else update(mid + 1, r, tree[x].r, tree[y].r, pos);
54 }
55
56 int query(int l, int r, int x, int y, int k) {
57 if(l == r) return l;
58 int mid = (l + r) >> 1;
59 int sum = tree[tree[y].l].sum - tree[tree[x].l].sum;
60 if(sum >= k) return query(l, mid, tree[x].l, tree[y].l, k);
61 else return query(mid + 1, r, tree[x].r, tree[y].r, k - sum);
62 }
63
64 int main() {
65 while(~scanf("%d%d", &n, &q)) {
66 cnt = 0;
67 for(int i = 1; i <= n; i++) scanf("%d", &a[i]), update(1, n, root[i], root[i-1], a[i]);
68 while(q--) {
69 scanf("%d%d", &x, &y);
70 int len = y - x + 1;
71 int ub = len, lb = 1, mid, ans = 1;
72 while(ub >= lb) {
73 mid = (ub + lb) >> 1;
74 int pp = query(1, n, root[x-1], root[y], len - mid + 1);
75 if(pp >= mid) {
76 lb = mid + 1;
77 ans = max(ans, mid);
78 } else {
79 ub = mid - 1;
80 }
81 }
82 printf("%d\n", ans);
83 }
84 for(int i = 0; i <= cnt; i++) {
85 tree[i].sum = 0;
86 }
87 }
88 return 0;
89 }
F题：
题意：
给你n对三元组和排序原则，要你输出排序后的编号。
思路：
直接sort即可，不过需要注意精度，我们可以稍加化简排除精度问题，去年比赛的时候因为精度贡献了两发罚时，最后同队学长提醒一句就A了。
代码实现如下：
1 #include <set>
2 #include <map>
3 #include <deque>
4 #include <queue>
5 #include <stack>
6 #include <cmath>
7 #include <ctime>
8 #include <bitset>
9 #include <cstdio>
10 #include <string>
11 #include <vector>
12 #include <cstdlib>
13 #include <cstring>
14 #include <iostream>
15 #include <algorithm>
16 using namespace std;
17
18 typedef long long LL;
19 typedef pair<LL, LL> pLL;
20 typedef pair<LL, int> pLi;
21 typedef pair<int, LL> pil;;
22 typedef pair<int, int> pii;
23 typedef unsigned long long uLL;
24
25 #define lson rt<<1
26 #define rson rt<<1|1
27 #define lowbit(x) x&(-x)
28 #define name2str(name) (#name)
29 #define bug printf("*********\n")
30 #define debug(x) cout<<#x"=["<<x<<"]" <<endl
31 #define FIN freopen("D://code//in.txt","r",stdin)
32 #define IO ios::sync_with_stdio(false),cin.tie(0)
33
34 const double eps = 1e-8;
35 const int mod = 1000000007;
36 const int maxn = 1e3 + 7;
37 const double pi = acos(-1);
38 const int inf = 0x3f3f3f3f;
39 const LL INF = 0x3f3f3f3f3f3f3f3fLL;
40
41 int n;
42
43 struct node {
44 int id;
45 LL a, b, c;
46 bool operator < (const node& x) const {
47 return (a + b) * x.c == (x.a + x.b) * c ? id < x.id : (a + b) * x.c < (x.a + x.b) * c;
48 }
49 }num[maxn];
50
51 int main(){
52 while(~scanf("%d", &n)) {
53 for(int i = 1; i <= n; i++) {
54 num[i].id = i;
55 scanf("%lld%lld%lld", &num[i].a, &num[i].b, &num[i].c);
56 }
57 sort(num + 1, num + n + 1);
58 for(int i = 1; i <= n; i++) {
59 printf("%d%c", num[i].id, i == n ? ‘\n‘ : ‘ ‘);
60 }
61 }
62 return 0;
63 }
G题：
题意：
给你两个字符串，然后其中aa、bb和abab可以消除或者在任意地方添加，然后问你s可以经过任意次操作变成t不？
思路：
这题的样例给的十分贴心，从第一个样例可以看出a和b可以互换位置，第二个样例可以推出a和b不能与c换位置，故对于s和t我们以c为分界，查看每一段内s的ab的奇偶是否等于t对应段的奇偶。
代码实现如下：
1 #include <set>
2 #include <map>
3 #include <deque>
4 #include <queue>
5 #include <stack>
6 #include <cmath>
7 #include <ctime>
8 #include <bitset>
9 #include <cstdio>
10 #include <string>
11 #include <vector>
12 #include <cstdlib>
13 #include <cstring>
14 #include <iostream>
15 #include <algorithm>
16 using namespace std;
17
18 typedef long long LL;
19 typedef pair<LL, LL> pLL;
20 typedef pair<LL, int> pLi;
21 typedef pair<int, LL> pil;;
22 typedef pair<int, int> pii;
23 typedef unsigned long long uLL;
24
25 #define lson rt<<1
26 #define rson rt<<1|1
27 #define lowbit(x) x&(-x)
28 #define name2str(name) (#name)
29 #define bug printf("*********\n")
30 #define debug(x) cout<<#x"=["<<x<<"]" <<endl
31 #define FIN freopen("D://code//in.txt","r",stdin)
32 #define IO ios::sync_with_stdio(false),cin.tie(0)
33
34 const double eps = 1e-8;
35 const int mod = 1000000007;
36 const int maxn = 1e4 + 7;
37 const double pi = acos(-1);
38 const int inf = 0x3f3f3f3f;
39 const LL INF = 0x3f3f3f3f3f3f3f3fLL;
40
41 string ans1, ans2;
42 char s[maxn], t[maxn];
43
44 int main(){
45 while(~scanf("%s%s", s, t)) {
46 int n = strlen(s), m = strlen(t);
47 ans1.clear(); ans2.clear();
48 int num1 = 0, num2 = 0;
49 for(int i = 0; i < n; i++) {
50 if(s[i] == ‘a‘) {
51 num1 = (num1 + 1) % 2;
52 } else if(s[i] == ‘b‘) {
53 num2 = (num2 + 1) % 2;
54 } else {
55 if(num1) ans1 += ‘a‘;
56 if(num2) ans1 += ‘b‘;
57 ans1 += ‘c‘;
58 num1 = num2 = 0;
59 }
60 }
61 if(num1) ans1 += ‘a‘;
62 if(num2) ans1 += ‘b‘;
63 num1 = num2 = 0;
64 for(int i = 0; i < m; i++) {
65 if(t[i] == ‘a‘) {
66 num1 = (num1 + 1) % 2;
67 } else if(t[i] == ‘b‘) {
68 num2 = (num2 + 1) % 2;
69 } else {
70 if(num1) ans2 += ‘a‘;
71 if(num2) ans2 += ‘b‘;
72 ans2 += ‘c‘;
73 num1 = num2 = 0;
74 }
75 }
76 if(num1) ans2 += ‘a‘;
77 if(num2) ans2 += ‘b‘;
78 if(ans1 == ans2) puts("Yes");
79 else puts("No");
80 }
81 return 0;
82 }
K题：
题意：
在[a,b]内选择一个数x，同时在[c,d]中选择一个数y，使得x*y%2018=0的方案数。
思路：
枚举+容斥。
代码实现如下：
1 #include <set>
2 #include <map>
3 #include <deque>
4 #include <queue>
5 #include <stack>
6 #include <cmath>
7 #include <ctime>
8 #include <bitset>
9 #include <cstdio>
10 #include <string>
11 #include <vector>
12 #include <cstdlib>
13 #include <cstring>
14 #include <iostream>
15 #include <algorithm>
16 using namespace std;
17
18 typedef long long LL;
19 typedef pair<LL, LL> pLL;
20 typedef pair<LL, int> pLi;
21 typedef pair<int, LL> pil;;
22 typedef pair<int, int> pii;
23 typedef unsigned long long uLL;
24
25 #define lson rt<<1
26 #define rson rt<<1|1
27 #define lowbit(x) x&(-x)
28 #define name2str(name) (#name)
29 #define bug printf("*********\n")
30 #define debug(x) cout<<#x"=["<<x<<"]" <<endl
31 #define FIN freopen("D://code//in.txt","r",stdin)
32 #define IO ios::sync_with_stdio(false),cin.tie(0)
33
34 const double eps = 1e-8;
35 const int mod = 1000000007;
36 const int maxn = 2e5 + 7;
37 const double pi = acos(-1);
38 const int inf = 0x3f3f3f3f;
39 const LL INF = 0x3f3f3f3f3f3f3f3fLL;
40
41 int a, b, c, d;
42
43 int main(){
44 while(~scanf("%d%d%d%d", &a, &b, &c, &d)) {
45 LL ans = 1LL * (b - a + 1) * (d / 2018 - (c - 1) / 2018);
46 ans += 1LL * (b / 2 - (a - 1) / 2 - (b / 2018 - (a - 1) / 2018)) * (d / 1009 - (c - 1) / 1009 - (d / 2018 - (c - 1) / 2018));
47 ans += 1LL * (b / 1009 - (a - 1) / 1009 - (b / 2018 - (a - 1) / 2018)) * (d / 2 - (c - 1) / 2 - (d / 2018 - (c - 1) / 2018));
48 ans += 1LL * (b / 2018 - (a - 1) / 2018) * (d - c + 1 - (d / 2018 - (c - 1) / 2018));
49 printf("%lld\n", ans);
50 }
51 return 0;
52 }
去年和学长、同届一个OI大佬组队，全程被带飞，全程就贡献了一个F，希望今年湘潭邀请赛自己能够carry，夺一块好牌子吧~
原文地址：https://www.cnblogs.com/Dillonh/p/10428566.html
时间： 2024-10-09 02:30:03