Nick has n bottles of soda left after his birthday. Each bottle is described by two values: remaining amount of soda ai and bottle volume bi (ai ≤ bi).
Nick has decided to pour all remaining soda into minimal number of bottles, moreover he has to do it as soon as possible. Nick spends x seconds to pour x units of soda from one bottle to another.
Nick asks you to help him to determine k — the minimal number of bottles to store all remaining soda and t — the minimal time to pour soda into k bottles. A bottle can't store more soda than its volume. All remaining soda should be saved.
The first line contains positive integer n (1 ≤ n ≤ 100) — the number of bottles.
The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is the amount of soda remaining in the i-th bottle.
The third line contains n positive integers b1, b2, ..., bn (1 ≤ bi ≤ 100), where bi is the volume of the i-th bottle.
It is guaranteed that ai ≤ bi for any i.
The only line should contain two integers k and t, where k is the minimal number of bottles that can store all the soda and t is the minimal time to pour the soda into k bottles.
4
3 3 4 3
4 7 6 5
2 6
2
1 1
100 100
1 1
5
10 30 5 6 24
10 41 7 8 24
3 11
In the first example Nick can pour soda from the first bottle to the second bottle. It will take 3 seconds. After it the second bottle will contain 3 + 3 = 6 units of soda. Then he can pour soda from the fourth bottle to the second bottle and to the third bottle: one unit to the second and two units to the third. It will take 1 + 2 = 3 seconds. So, all the soda will be in two bottles and he will spend 3 + 3 = 6 seconds to do it.
题意：给你n个杯子 给出n个杯子的初始水的体积以及杯子的体积 现在倒水 使得用最少个数的杯子装所有的水 并且在倒水的过程中 使得倒出的水尽量的少
题解：爆搜TLE 背包处理 orzzz 太菜了
dp[i][j]表示用i个杯子容量为j最多能装多少水 注意这个j的范围为杯子的体积和 （可能所选的i个杯子 就算装有全部的水 也不会装满）
1 /****************************** 2 code by drizzle 3 blog: www.cnblogs.com/hsd-/ 4 ^ ^ ^ ^ 5 O O 6 ******************************/ 7 #include<bits/stdc++.h> 8 #include<map> 9 #include<set> 10 #include<cmath> 11 #include<queue> 12 #include<bitset> 13 #include<math.h> 14 #include<vector> 15 #include<string> 16 #include<stdio.h> 17 #include<cstring> 18 #include<iostream> 19 #include<algorithm> 20 #pragma comment(linker, "/STACK:102400000,102400000") 21 using namespace std; 22 #define A first 23 #define B second 24 const int mod=1000000007; 25 const int MOD1=1000000007; 26 const int MOD2=1000000009; 27 const double EPS=0.00000001; 28 typedef __int64 ll; 29 const ll MOD=1000000007; 30 const int INF=1000000010; 31 const ll MAX=1ll<<55; 32 const double eps=1e-5; 33 const double inf=~0u>>1; 34 const double pi=acos(-1.0); 35 typedef double db; 36 typedef unsigned int uint; 37 typedef unsigned long long ull; 38 int n; 39 struct node 40 { 41 int a,b; 42 }N[105]; 43 int dp[105][10005]; 44 bool cmp(struct node aa,struct node bb) 45 { 46 if(aa.b>bb.b) 47 return true; 48 if(aa.b==bb.b) 49 { 50 if(aa.a>bb.a) 51 return true; 52 } 53 return false; 54 } 55 int main() 56 { 57 scanf("%d",&n); 58 int m=0; 59 int mm; 60 int mm2=0; 61 for(int i=1;i<=n;i++) 62 { 63 scanf("%d",&N[i].a); 64 m+=N[i].a; 65 } 66 mm=m; 67 for(int i=1;i<=n;i++) 68 { 69 scanf("%d",&N[i].b); 70 mm2+=N[i].b; 71 } 72 sort(N+1,N+1+n,cmp); 73 int num=0; 74 while(1) 75 { 76 m-=N[++num].b; 77 if(m<=0) 78 break; 79 } 80 memset(dp,0,sizeof(dp)); 81 for(int i=0;i<=num;i++) 82 for(int j=0;j<=mm2;j++) 83 dp[i][j]=-INF; 84 dp[0][0]=0; 85 for(int j=1;j<=n;j++) 86 { 87 for(int k=mm2;k>=N[j].b;k--) 88 { 89 for(int l=1;l<=num;l++){ 90 dp[l][k]=max(dp[l][k],dp[l-1][k-N[j].b]+N[j].a); 91 } 92 } 93 } 94 int ans=0; 95 for(int i=mm2;i>=mm;i--) 96 if(dp[num][i]) 97 ans=max(ans,dp[num][i]); 98 printf("%d %d\n",num,mm-ans); 99 return 0; 100 } 101 /* 102 10 103 18 42 5 1 26 8 40 34 8 29 104 18 71 21 67 38 13 99 37 47 76 105 3 100 106 */
