Haut OJ 1316: sister choice buys candy III

Problem description ：

Choice The schoolgirl looks at so many sweets in the candy store , She thought it would be great if she could take them all home . Now there are n Grow candy （1-n Number ）, Each kind of candy has mi individual （mi On behalf of the i The number of candies ）,Choice The student sister only has two paper bags , Each paper bag can hold at most w A candy .Choice The schoolgirl doesn't want a paper bag mixed with a variety of sweets （ One paper bag at a time can only contain one kind of candy ）, that Choice How many times does the schoolgirl have to take all these sweets home at least ?

Input ：

Enter two integers on the first line n,w（1 <= n ,w<= 1e4）, Space off

Next line n Space separated integers mi, It means the first one i The number of candies .(1<= mi <= 1e3）.

Output ：

How many times should I take the output at least , The answer is on one line .

The sample input ：

6 5 
3 2 6 5 4 4

Sample output ：

4

---

Cause analysis ：

Calculate the total number of bags needed , Number of bags /2 According to the circumstances , Because because / The result is an integer , That is, odd numbers will be rounded down , One less time , So you can Odd number Even numbers Merge     (ans+ 1) / 2    for example 7/2=3 , Mathematically, it should be equal to 3.5, The actual need 4 Time , even numbers 8 need 4 Time , Add 1 There are still four times       Actually ans/2  + 0.5   +0.5 It is equivalent to rounding

in the future Practical application   /2  Remember to think about Odd and even cases , Whether it will have an impact  

---

Solution ：

#include <stdio.h>

int main()
{
    int n, w, t;
    int ans = 0;
    scanf("%d%d", &n, &w);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &t);
        ans += t / w;
        if(t % w)
            ans ++;
    }
    printf("%d\n", (ans+ 1) / 2);
    return 0;
}