HDU4519

一种比较挫的写法

 1 /*

 2 模拟

 3 */

 4 #include<stdio.h>

 5 #include<string.h>

 6 #include<stdlib.h>

 7 #include<algorithm>

 8 #include<iostream>

 9 #include<queue>

10 #include<stack>

11 #include<math.h>

12 #include<map>

13 using namespace std;

14 const int maxn = 105;

15 

16 int cmp( int a,int b ){

17     return a>b;

18 }

19 

20 int main(){

21     int ca;

22     scanf("%d",&ca);

23     while( ca-- ){

24         int n,m,k;

25         scanf("%d%d%d",&n,&k,&m);

26         int a[ maxn ];

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

28             a[i] = k;

29         int ans = 0;

30         while( 1 ){

31             sort( a,a+n,cmp );

32             if( a[0]<=0 ) break;

33             for( int i=0;a[i]>0&&i<min(n,m);i++ ) 

34                 a[ i ]--;

35             ans++;

36         }

37         printf("%d\n",ans);

38     }

39     return 0;

40 }

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另外一种方法

 1 /*

 2 模拟

 3 */

 4 #include<stdio.h>

 5 #include<string.h>

 6 #include<stdlib.h>

 7 #include<algorithm>

 8 #include<iostream>

 9 #include<queue>

10 #include<stack>

11 #include<math.h>

12 #include<map>

13 using namespace std;

14 const int maxn = 105;

15 const double eps = 0.99999999;

16 

17 int main(){

18     int ca;

19     scanf("%d",&ca);

20     while( ca-- ){

21         int n,m,k;

22         scanf("%d%d%d",&n,&k,&m);

23         if( m>=n ){

24             printf("%d\n",k);

25             continue;

26         }

27         if( (n*k)%m==0 ) printf("%d\n",n*k/m);

28         else printf("%d\n",n*k/m+1);

29     }

30     return 0;

31 }

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