题目描述
Ibis is fighting with a monster.
The health of the monster is H.
Ibis can cast N kinds of spells. Casting the i-th spell decreases the monster’s health by Ai, at the cost of Bi Magic Points.
The same spell can be cast multiple times. There is no way other than spells to decrease the monster’s health.
Ibis wins when the health of the monster becomes 0 or below.
Find the minimum total Magic Points that have to be consumed before winning.
Constraints
·1≤H≤104
·1≤N≤103
·1≤Ai≤104
·1≤Bi≤104
·All values in input are integers.
输入
Input is given from Standard Input in the following format:
H N A1 B1 : AN BN
输出
Print the minimum total Magic Points that have to be consumed before winning.
样例输入
【样例1】 9 3 8 3 4 2 2 1 【样例2】 100 6 1 1 2 3 3 9 4 27 5 81 6 243 【样例3】 9999 10 540 7550 691 9680 700 9790 510 7150 415 5818 551 7712 587 8227 619 8671 588 8228 176 2461
样例输出
【样例1】 4 【样例2】 100 【样例3】 139815
提示
样例1解释
First, let us cast the first spell to decrease the monster’s health by 8, at the cost of 3 Magic Points. The monster’s health is now 1.
Then, cast the third spell to decrease the monster’s health by 2, at the cost of 1 Magic Point. The monster’s health is now −1.
In this way, we can win at the total cost of 4 Magic Points.
样例2解释
It is optimal to cast the first spell 100 times.
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math") #pragma GCC optimize("Ofast") #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") #pragma comment(linker, "/stack:200000000") #pragma GCC optimize (2) #pragma G++ optimize (2) #include <bits/stdc++.h> #include <algorithm> #include <map> #include <queue> #include <set> #include <stack> #include <string> #include <vector> using namespace std; #define wuyt main typedef long long ll; #define HEAP(...) priority_queue<__VA_ARGS__ > #define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > > template<class T> inline T min(T &x,const T &y){return x>y?y:x;} template<class T> inline T max(T &x,const T &y){return x<y?y:x;} ///#define getchar()(p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++) ///char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf; ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar(); if(c == '-')Nig = -1,c = getchar(); while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar(); return Nig*x;} #define read read() const ll inf = 1e15; const int maxn = 2e5 + 7; const int mod = 1e9 + 7; #define start int wuyt() int num[maxn]; ll ans; int main() { int h=read,n=read; vector<int> dp(h+1, mod); dp[0]=0; for(int i=0;i<n;i++) { int a=read,b=read; for(int j=0;j<h;j++) { int temp=min(j+a,h); dp[temp] = min(dp[temp],dp[j]+b); } } printf("%d\n",dp[h]); return 0; }
