CodeForce 517 Div 2. C Cram Time

http://codeforces.com/contest/1072/problem/C

C. Cram Time

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

In a galaxy far, far away Lesha the student has just got to know that he has an exam in two days. As always, he hasn't attended any single class during the previous year, so he decided to spend the remaining time wisely.

Lesha knows that today he can study for at most $\mathrm{aa hours,\; and\; he\; will\; have bb hours\; to\; study\; tomorrow.\; Note\; that\; it\; is\; possible\; that\; on\; his\; planet\; there\; are\; more\; hours\; in\; a\; day\; than\; on\; Earth.\; Lesha\; knows\; that\; the\; quality\; of\; his\; knowledge\; will\; only\; depend\; on\; the\; number\; of\; lecture\; notes\; he\; will\; read.\; He\; has\; access\; to\; an\; infinite\; number\; of\; notes\; that\; are\; enumerated\; with\; positive\; integers,\; but\; he\; knows\; that\; he\; can\; read\; the\; first\; note\; in\; one\; hour,\; the\; second\; note\; in\; two\; hours\; and\; so\; on.\; In\; other\; words,\; Lesha\; can\; read\; the\; note\; with\; number kk in kkhours.\; Lesha\; can\; read\; the\; notes\; in\; arbitrary\; order,\; however,\; he\; can\text{'}t\; start\; reading\; a\; note\; in\; the\; first\; day\; and\; finish\; its\; reading\; in\; the\; second\; day.}$

Thus, the student has to fully read several lecture notes today, spending at most $\mathrm{aa hours\; in\; total,\; and\; fully\; read\; several\; lecture\; notes\; tomorrow,\; spending\; at\; most bb hours\; in\; total.\; What\; is\; the\; maximum\; number\; of\; notes\; Lesha\; can\; read\; in\; the\; remaining\; time?\; Which\; notes\; should\; he\; read\; in\; the\; first\; day,\; and\; which —\; in\; the\; second?}$

Input

The only line of input contains two integers $\mathrm{aa and bb (0\le a,b\le 1090\le a,b\le 109) —\; the\; number\; of\; hours\; Lesha\; has\; today\; and\; the\; number\; of\; hours\; Lesha\; has\; tomorrow.}$

Output

In the first line print a single integer $\mathrm{nn (0\le n\le a0\le n\le a) —\; the\; number\; of\; lecture\; notes\; Lesha\; has\; to\; read\; in\; the\; first\; day.\; In\; the\; second\; line\; print nn distinct\; integers p1,p2,\dots ,pnp1,p2,\dots ,pn (1\le pi\le a1\le pi\le a),\; the\; sum\; of\; all pipi should\; not\; exceed aa.}$

In the third line print a single integer $\mathrm{mm (0\le m\le b0\le m\le b) —\; the\; number\; of\; lecture\; notes\; Lesha\; has\; to\; read\; in\; the\; second\; day.\; In\; the\; fourth\; line\; print mm distinct\; integers q1,q2,\dots ,qmq1,q2,\dots ,qm (1\le qi\le b1\le qi\le b),\; the\; sum\; of\; all qiqi should\; not\; exceed bb.}$

All integers ${\mathrm{pipi and qiqi should\; be\; distinct. The\; sum n+mn+m should\; be\; largest\; possible.}}^{}$

Examples

input

Copy

3 3

output

Copy

1
3 
2
2 1 

input

9 12

output

2
3 6
4
1 2 4 5

Note

In the first example Lesha can read the third note in $\mathrm{33 hours\; in\; the\; first\; day,\; and\; the\; first\; and\; the\; second\; notes\; in\; one\; and\; two\; hours\; correspondingly\; in\; the\; second\; day,\; spending 33 hours\; as\; well.\; Note\; that\; Lesha\; can\; make\; it\; the\; other\; way\; round,\; reading\; the\; first\; and\; the\; second\; notes\; in\; the\; first\; day\; and\; the\; third\; note\; in\; the\; second\; day.}$

In the second example Lesha should read the third and the sixth notes in the first day, spending $\mathrm{99 hours\; in\; total.\; In\; the\; second\; day\; Lesha\; should\; read\; the\; first,\; second\; fourth\; and\; fifth\; notes,\; spending 1212 hours\; in\; total.}$

 

题意：

　有无数多个lecture note，标号为1，2，3，4，5... 他们复习所要花费的时间和标号大小相同。分两天复习这些lecture。每天的复习小时不一样，求分配使得复习的lecture最多。

思路：

首先a,b分开装东西，不会比他们合在一起装东西装的多。（反证法）

其次，如果从大到小枚举可行的lecture note，如果每次先考虑a，那么a最后肯定可以被装满。

所以，剩下的lecture note，给b，b一定装的下所有。

代码：

 1 #include <iostream>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <cmath>
 5 #include <cstdio>
 6 #include <vector>
 7 #include <queue>
 8 #include <set>
 9 #include <map>
10 #include <stack>
11 #define ll long long
12 #define local
13 
14 using namespace std;
15 
16 const int MOD = 1e9+7;
17 const int inf = 0x3f3f3f3f;
18 const double PI = acos(-1.0);
19 const int maxn = (1e5+10);
20 const int maxedge = 100*100;
21 ll a, b;
22 
23 int main() {
24 #ifdef local
25     if(freopen("/Users/Andrew/Desktop/data.txt", "r", stdin) == NULL) printf("can't open this file!\n");
26 #endif
27     
28     scanf("%lld%lld", &a, &b);
29     vector <ll> ansa, ansb;
30     ll x = 1;
31     for (; 1LL*x*(x+1)/2 <= a+b; ++x) ;
32     x--;
33     for (; x > 0; x--) {
34         if (a >= x) {
35             ansa.push_back(x);
36             a -= x;
37         } else if (b >= x) {
38             ansb.push_back(x);
39             b -= x;
40         }
41     }
42     printf("%d\n", int(ansa.size()));
43     for (int i = 0; i < ansa.size(); ++i) {
44         printf("%lld", ansa[i]);
45         if (i != ansa.size()-1) printf(" ");
46     }
47     cout << endl;
48     printf("%d\n", int(ansb.size()));
49     for (int i = 0; i < ansb.size(); ++i) {
50         printf("%lld", ansb[i]);
51         if (i != ansb.size()-1) printf(" ");
52     }
53     cout << endl;
54     
55 #ifdef local
56     fclose(stdin);
57 #endif
58     return 0;
59 }

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转载于:https://www.cnblogs.com/lecoz/p/9833990.html