Jafar has n cans of cola. Each can is described by two integers: remaining volume of cola ai and can's capacity bi (ai ≤ bi).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!
Input
The first line of the input contains one integer n (2 ≤ n ≤ 100 000) — number of cola cans.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 10^9) — volume of remaining cola in cans.
The third line contains n space-separated integers that b1, b2, ..., bn (ai ≤ bi ≤ 10^9) — capacities of the cans.
Output
Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
2 3 5 3 6
Output
YES
Input
3 6 8 9 6 10 12
Output
NO
Input
5 0 0 5 0 0 1 1 8 10 5
Output
YES
Input
4 4 1 0 3 5 2 2 3
Output
YES
Note
In the first sample, there are already 2 cans, so the answer is "YES".
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
int main()
{
int n, a[100010], b[100010], i, j, t, sum;
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
scanf("%d", &n);
sum = 0;
for(i = 0; i < n; i++)
{
scanf("%d", &a[i]);
sum += a[i];
}
for(i = 0; i < n; i++)
{
scanf("%d", &b[i]);
}
for(i = 0; i < n; i++)
{
for(j = 0; j < n - 1 - i; j++)
{
if(b[j] < b[j + 1])
{
t = b[j];
b[j] = b[j + 1];
b[j + 1] = t;
}
}
}
if(sum <= (b[0] + b[1])) printf("YES\n");
else printf("NO\n");
return 0;
}上方代码 time limit exceeded
后面我把冒泡排序改为选择排序，并且只排除两个最大的数，但是变成了 wrong answer
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
int main()
{
int n, a[100010], b[100010], i, j, t, sum, k, max;
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
scanf("%d", &n);
sum = 0;
for(i = 0; i < n; i++)
{
scanf("%d", &a[i]);
sum += a[i];
}
for(i = 0; i < n; i++)
{
scanf("%d", &b[i]);
}
for(i = 0; i < 2; i++)
{
k = i;
max = k;
for(j = k; j < n; j++)
{
if(b[max] < b[j]) max = j;
}
if(max != k)
{
t = b[k];
b[k] = b[max];
b[max] = t;
}
}
if(sum <= (b[0] + b[1])) printf("YES\n");
else printf("NO\n");
return 0;
}
改正点：
1. 题目所给数过大，n大小为1e5，sum大小可能为1e10，而int范围是2^32=4e9， long long int 范围是1.8e19，此处sum的定义应定义为long long int
2. 此题不需要排序，直接找出最大的两个数就好，只需要循环2次，比排序耗时少很多
改正代码：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
long long int a[100010], b[100010];//数组开在main外面比较稳定
int main()
{
int n, i, max1, k, max2;
long long int sum;//这道题的数据量太大，sum定义为int会出问题
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
scanf("%d", &n);
sum = 0;
for(i = 0; i < n; i++)
{
scanf("%lld", &a[i]);
sum += a[i];//计算可乐的总量
}
max1 = 0;
for(i = 0; i < n; i++)
{
scanf("%lld", &b[i]);
if(max1 < b[i])
{
max1 = b[i];
k = i;
}
}//找到第一大容量的瓶子,并记录其下标
max2 = 0;
for(i = 0; i < n; i++)
{
if(max2 < b[i] && i != k) max2 = b[i];
}//找到第二大容量的瓶子，并且这个数不是第一个数
if(sum <= (max1 + max2)) printf("YES\n");
else printf("NO\n");
return 0;
}