UESTC567 Journey(枚举)

题解详见http://www.desgard.com/learning/2015/03/25/UESTC_567/

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<cstdlib>

 4 #include<cstring>

 5 #include<string>

 6 #include<queue>

 7 #include<algorithm>

 8 #include<map>

 9 #include<iomanip>

10 #include<climits>

11 #include<string.h>

12 #include<numeric>

13 #include<cmath>

14 #include<stdlib.h>

15 #include<vector>

16 #include<stack>

17 #include<set>

18 #define INF 1e7

19 #define MAXN 100010

20 #define maxn 1000010

21 #define Mod 1000007

22 #define N 1010

23 using namespace std;

24 typedef long long LL;

25 

26 int T;

27 int kase = 1, n;

28 int e[2010];

29 LL res, sum;

30 void run()

31 {

32     sum = 0;

33     scanf("%d", &n);

34     for (int i = 1; i <= n; ++i){

35         scanf("%d", &e[i]);

36         e[i + n] = e[i];

37         sum += e[i];

38     }

39     e[0] = e[n];

40     e[2 * n + 1] = e[1];

41     printf("Case #%d:", kase++);

42     for (int i = 1; i <= n; ++i) {

43         LL ans = sum - e[i] < sum - e[i - 1] ? sum - e[i] : sum - e[i - 1];

44         LL temp = 0;

45         for (int c = 0; c < n; ++c) {

46             temp += e[i + c];

47             ans = min(ans, temp + sum - e[i + c + 1]);

48         }

49         temp = 0;

50         for (int c = 1; c < n; ++c) {

51             int j = i - c;

52             if (j <= 0) j += n;

53             temp += e[j];

54             ans = min(ans, temp + sum - e[j - 1]);

55         }

56         printf(" %lld", ans);

57     }

58     puts("");

59 }

60 

61 int main()

62 {

63     cin >> T;

64     while (T--)

65         run();

66     return 0;

67 }