Battle ShipsTime Limit: 2 Seconds Memory Limit: 65536 KB
Battle Ships is a new game which is similar to Star Craft. In this game, the enemy builds a defense tower, which hasL longevity. The player has a military factory, which can produce N kinds of battle ships. The factory takes ti seconds to produce thei-th battle ship and this battle ship can make the tower loss li longevity every second when it has been produced. If the longevity of the tower lower than or equal to 0, the player wins. Notice that at each time, the factory can choose only one kind of battle ships to produce or do nothing. And producing more than one battle ships of the same kind is acceptable.
Your job is to find out the minimum time the player should spend to win the game.
Input
There are multiple test cases. The first line of each case contains two integers N(1 ≤ N ≤ 30) andL(1 ≤ L ≤ 330), N is the number of the kinds of Battle Ships,L is the longevity of the Defense Tower. Then the following N lines, each line contains two integerst i(1 ≤ t i ≤ 20) andli(1 ≤ li ≤ 330) indicating the produce time and the lethality of the i-th kind Battle Ships.
Output
Output one line for each test case. An integer indicating the minimum time the player should spend to win the game.
Sample Input1 11 12 101 12 53 1001 103 2010 100Sample Output245
题意：有n种船给出造出每种船的时间和该船一秒可以打多少滴血，，给出L总血量，问打到L滴血最少需要多少秒。
设dp[i]代表第i秒最多可以打掉多少滴血那么答案就是从第0秒开始枚举找到第一个大于等于L的就结束。
dp[j+t[i]] = max{dp[i+t[i]], dp[j] + l[i]*j}
#include <algorithm>#include <cstdio>#include <vector>#include <cmath>#include <queue>#include <set>#include <map>#include <cstring>#include <cstdlib>#include <iostream>//#include <bits/stdc++.h>#define MAX 0x3f3f3f3f#define N 100005#define M 200005#define mod 1000000007#define lson o<<1, l, m#define rson o<<1|1, m+1, rtypedef long long LL;using namespace std;const double pi = acos(-1.0);int n, m;int t[40], l[40], dp[40000];int main(){//freopen("in.txt","r",stdin);while(~scanf("%d%d", &n, &m)) {for(int i = 0; i < n; i++) {scanf("%d%d", &t[i], &l[i]);}memset(dp, 0, sizeof(dp));for(int i = 0; i < n; i++) {for(int j = 0; j <= 340; j++) {dp[ t[i] + j ] = max(dp[ t[i] + j ], dp[j] + j*l[i]);}}for(int i = 0; i < 400; i++) {if(dp[i] >= m) {printf("%d\n", i);break;}}}return 0;}
会让你的心态更平和更坦然，也会让你心无旁骛，更会让你的心灵得到解脱和抚慰。
