hdu2124 Repair the Wall

没多大意思的水题，题意是说修补屋子的1*n矩形漏洞，给定一些木块用最少的木块修补好裂缝，可使用锯子裁掉多余部分，应该算是简单的贪心算法，从大到小排序后累加，大于就可以跳出循环。。。。

附代码如下：

# include <iostream>

using namespace std;

__int64 a[650];

int cmp(int a, int b)

{

return a>b;

}

int main()

{

__int64 Lack, n;

while (scanf("%I64d %I64d", &Lack, &n) != EOF){



for (int i = 0; i < n; i ++){

scanf("%I64d", &a[i]);

}

sort(a, a+n, cmp);

__int64 sum = 0;

int check = 0;

int counter = 0;

for (int i = 0; i < n; i ++){

if (sum >= Lack){

check = 1;

break;

}

else {

sum += a[i];

counter ++;

}

}

if (sum >= Lack)check = 1;

if (check)printf("%d/n", counter);

else printf("impossible/n");

}

return 0;

}