The cows are going to space! They plan to achieve orbit by building a sort of space elevator: a giant tower of blocks. They have K (1 <= K <= 400) different types of blocks with which to build the tower. Each block of type i has height h_i (1 <= h_i <= 100) and is available in quantity c_i (1 <= c_i <= 10). Due to possible damage caused by cosmic rays, no part of a block of type i can exceed a maximum altitude a_i (1 <= a_i <= 40000).

Help the cows build the tallest space elevator possible by stacking blocks on top of each other according to the rules.

* Line 1: A single integer, K

* Lines 2..K+1: Each line contains three space-separated integers: h_i, a_i, and c_i. Line i+1 describes block type i.

* Line 1: A single integer H, the maximum height of a tower that can be built

3
7 40 3
5 23 8
2 52 6

48

OUTPUT DETAILS:

From the bottom: 3 blocks of type 2, below 3 of type 1, below 6 of type 3. Stacking 4 blocks of type 2 and 3 of type 1 is not legal, since the top of the last type 1 block would exceed height 40.

#include <iostream>
#include <algorithm>
using namespace std;
int K;
const int MAXN = 40010;
bool dp[40010];
int h[MAXN], a[MAXN];
int V;struct block {int h, a, c;bool operator<(const block &other) const {return this->a < other.a;}
};block blocks[MAXN];
int pre_process() {sort(blocks, blocks+K);int len = 0;for(int i = 0; i < K; i ++) {int rem = blocks[i].c;int j = 0;// 预处理, while(rem) {if(rem >= (1<<j)) {h[len] = (1<<j)*blocks[i].h;a[len] = blocks[i].a;rem -= (1<<j);j++;if(h[len] > a[len])continue;len++;}else{h[len] = rem*blocks[i].h;a[len] = blocks[i].a;len++;rem = 0;}}}return len;
}int solve_dp() {int len = pre_process();// 01背包memset(dp, 0, sizeof(dp));dp[0] = 1;for(int i = 0; i < len; i ++) {for(int v = V; v >= h[i]; v--) {if(v > a[i]) continue;if(!dp[v] && dp[v-h[i]]) {dp[v] = true;//printf("dp[%d] = true\n", v);}}}for(int v = V; v >= 0; v --)if(dp[v])return v;
}
int main() {freopen("E:\\Copy\\ACM\\测试用例\\in.txt", "r", stdin);cin >> K;int h_i, a_i, c_i;V = 0;for(int i = 0; i < K; i ++) {scanf("%d%d%d", &blocks[i].h, &blocks[i].a, &blocks[i].c);V = min(V+blocks[i].h*blocks[i].c, 40000);}cout << solve_dp() << endl;return 0;
}