#include <bits/stdc++.h>
using namespace std;
#define f(n,m,i) for (int i = n;i <= m;i++)
#define fc(n,m,i) for (int i = n;i >= m;i--)
#define dbug(x) cerr<<(#x)<<':'<<x<<' ';
#define ent cerr<<endl;
inline void C(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cerr.tie(0);
}
double max(double x,double y){
	return x>y?x:y;
}
double min(double x,double y){
	return x<y?x:y;
}
long long n,maxt,km;
long long w[1005];
double t[1005];
long long wes[1005];
double dp[1005][1005];
int main(){
	C();
	cin >> maxt >> km >> n;
	int s;
	f(1,n,i){
		f(1,n,j){
			dp[i][j] = 1e15;
		}
	}
	f(1,n,i){
		cin >> w[i] >> s;
		t[i] = (km * 1.0) / s;
		wes[i] = wes[i - 1] + w[i];
		dp[i][i] = t[i];
		//dbug(t[i])
	}
	f(2,n,len){
		f(1,n - len + 1,i){
			int l = i,r = i + len - 1;
			
			if (wes[r] - wes[l - 1] <= maxt){
				f(l,r - 1,k){
					dp[l][r] = min(dp[l][r],max(dp[l][k],dp[k + 1][r]));
				}
			}
			else{
				f(l,r - 1,k){
					dp[l][r] = min(dp[l][r],dp[l][k] + dp[k + 1][r]);
				}
			}
		}
	}
	printf("%.1lf",dp[1][n] * 60);
	return 0;
}

本来写区间玩一玩，以为会T几个点，没想到过了，最慢的那个点都只有699ms。

本题或将多一个O（n^3）的做法