sgu 259

/*
  greedy method
  sort the leaflets by Li by descending order
  let sumTk = Ti + T2 + ..... Tk   
  answer = max ( sumTk + Lk),  k = 1...n


proofproof by contradiction :
  假设最优序列为 a = { 1, 2, 3 ....i, j ... n } (j = i + 1, Li < Lj )
  构造序列 b = { 1，2，3 .....j, i ....n }
  
  ans(a) = max {sumT(i-1)+Ti+Li, sumT(i-1)+Ti+Tj+Lj, sumTk+Lk }, k != i, j
  ans(b) = max {sumT(i-1)+Tj+Lj, sumT(i-1)+Tj+Ti+Li, sumTk+Lk }, k != i, j

  because Li < Lj, so   Ti+Tj+Lj > Tj+Ti+Li 
  and it is obvious that Ti+Tj+Lj > Tj+Lj
  
  we come to conclusion: ans(a) >= ans(b) , this means a is not the best ans.
  
  */

#include <stdio.h>
#include <stdlib.h>

#define SIZE 100

typedef struct {
    int T;
    int L;
}TL;

int cmp(const void* a, const void* b) {
    TL* pa = (TL*)a;
    TL* pb = (TL*)b;
    return (pb->L) -  (pa->L);
}

int main() {
    int n, i;
    TL tl[SIZE];
    scanf("%d", &n);
    for (i=0; i<n; ++i) {
        scanf("%d", &tl[i].T);
    }
    for (i=0; i<n; ++i) {
        scanf("%d", &tl[i].L);
    }

    qsort(tl, n, sizeof(TL), cmp);

    int s = 0;
    int e;
    int ans = 0;
    for (i=0; i<n; ++i) {
        s += tl[i].T;
        e = s + tl[i].L;
        if (e > ans) {
            ans = e;
        }
    }

    printf("%d\n", ans);
    return 0;
}