Problem Description
As one of the most powerful brushes in the world, zhx usually takes part in all kinds of contests.One day, zhx takes part in an contest. He found the contest very easy for him.There are
problems in the contest. He knows that he can solve the
problem in
i
units of time and he can get
i
points.As he is too powerful, the administrator is watching him. If he finishes the
problem before time
i
, he will be considered to cheat.zhx doesn’t really want to solve all these boring problems. He only wants to get no less than
points. You are supposed to tell him the minimal time he needs to spend while not being considered to cheat, or he is not able to get enough points.Note that zhx can solve only one problem at the same time. And if he starts, he would keep working on it until it is solved. And then he submits his code in no time.
Input
Multiply test cases(less than ). Seek as the end of the file.For each test, there are two integers
and
separated by a space. (
,
9
)Then come n lines which contain three integers
i
. (
9
)
Output
For each test case, output a single line indicating the answer. If zhx is able to get enough points, output the minimal time it takes. Otherwise, output a single line saying "zhx is naive!" (Do not output quotation marks).
Sample Input
1 31 4 73 64 1 86 8 101 5 22 710 4 110 2 3
Sample Output
78zhx is naive!
题意：
有n道题i题用时ti秒，得分vi，在li时间点之前不能做出来，而且一道题不能分开几次做（一旦开始做，必须在ti时间内把它做完）问得到w分的最小用时是多少
思路: 把题目按开始做的先后顺序排序，然后o1背包
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#include<set>#include<map>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define eps 1e-8typedef __int64 ll;#define fre(i,a,b) for(i = a; i <b; i++)#define free(i,b,a) for(i = b; i >= a;i–)#define mem(t, v) memset ((t) , v, sizeof(t))#define ssf(n)scanf("%s", n)#define sf(n)scanf("%d", &n)#define sff(a,b) scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pfprintf#define bugpf("Hi\n")using namespace std;#define INF 0x3f3f3f3f#define N 100005int dp[30*N];struct stud{ int t,l,va; bool operator < (const stud a) const {return l-t<a.l-a.t; }}f[40];int all,sum;int n,m;void solve(){int i,j;sort(f,f+n);int te;mem(dp,0);fre(i,0,n) free(te,all,f[i].l)if(te>=f[i].t)dp[te]=max(dp[te],dp[te-f[i].t]+f[i].va);fre(i,0,all) if(dp[i]>=m) break;pf("%d\n",i);}int main(){int i,j;while(~sff(n,m)){sum=all=0;int temp=0;fre(i,0,n) {sfff(f[i].t,f[i].va,f[i].l);all+=f[i].t;temp=max(temp,f[i].l);sum+=f[i].va; }if(sum<m) {pf("zhx is naive!\n");continue; }all=max(all,temp); solve();}return 0;}
，唯有斯人面上簌簌流下的，是点点无声无行的热泪。
