HDU4525+公式

 

一开始TLE了。。。

 1 /*

 2 模拟

 3 */

 4 #include<stdio.h>

 5 #include<math.h>

 6 const int maxn = 10005;

 7 typedef __int64 int64;

 8 int64 a[ maxn ];

 9 int main(){

10     int64 k;

11     int n,k1,k2 ;

12     int T;

13     scanf("%d",&T);

14     int ca = 1;

15     while( T-- ){

16         scanf("%d%d%d%I64d",&n,&k1,&k2,&k);

17         int64 sum = 0;

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

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

20             sum += a[ i ];

21         }

22         int64 cnt = 0;

23         int flag = -1;

24         while( 1 ){

25             if( sum>k ) {

26                 flag = 1;

27                 break;

28             }//ok

29             if( abs(k1+k2)<=1 ){

30                 flag = -1;

31                 break;

32             }

33             int64 temp = a[ 0 ];

34             sum = 0;

35             for( int i=0;i<n-1;i++ ){

36                 a[ i ] = k1*a[ i ]+k2*a[ i+1 ];

37                 sum += a[i];

38             }

39             a[ n-1 ] = k1*a[ n-1 ]+k2*temp;

40             sum += a[ n-1 ];

41             cnt++;

42         }

43         printf("Case #%d: ",ca++);

44         if( flag==1 ) printf("%I64d\n",cnt);

45         else printf("inf\n");

46     }

47     return 0;

48 }

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后来推出公式

sigma(Ai)*(K1+K2)^t,t表示天数

 1 /*

 2 模拟

 3 */

 4 #include<stdio.h>

 5 #include<math.h>

 6 const int maxn = 10005;

 7 typedef __int64 int64;

 8 int main(){

 9     int ca = 1;

10     int T;

11     scanf("%d",&T);

12     while( T-- ){

13         printf("Case #%d: ",ca++);

14         int n,k1,k2;

15         int64 k;

16         scanf("%d%d%d%I64d",&n,&k1,&k2,&k);

17         int64 sum = 0;

18         int temp;

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

20             scanf("%d",&temp);

21             sum += temp;

22         }

23         if( sum>k ){

24             printf("0\n");

25             continue;

26         }

27         if( abs(k1+k2)<=1 ){

28             printf("inf\n");

29             continue;

30         }

31         double ans1 = sum*1.0/100000.0;

32         double ans2 = (k)*1.0/100000.0;

33         int cnt = 0;

34         while( ans1<=ans2 ){

35             cnt++;

36             ans1 = ans1*(k1+k2);

37         }

38         printf("%d\n",cnt);

39     }

40     return 0;

41 }

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