http://acm.hdu.edu.cn/showproblem.php?pid=5280
One day, Xuejiejie gets an array $A$. Among all non-empty intervals of $A$, she wants to find the most beautiful one. She defines the beauty as the sum of the interval. The beauty of the interval---$[L,R]$ is calculated by this formula : beauty$(L,R) = A[L]+A[L+1]+……+A[R]$. The most beautiful interval is the one with maximum beauty.
But as is known to all, Xuejiejie is used to pursuing perfection. She wants to get a more beautiful interval. So she asks Mini-Sun for help. Mini-Sun is a magician, but he is busy reviewing calculus. So he tells Xuejiejie that he can just help her change one value of the element of $A$ to $P$ . Xuejiejie plans to come to see him in tomorrow morning.
Unluckily, Xuejiejie oversleeps. Now up to you to help her make the decision which one should be changed(You must change one element).
In the first line there is an integer $T$, indicates the number of test cases.
In each case, the first line contains two integers $n$ and $P$. $n$ means the number of elements of the array. $P$ means the value Mini-Sun can change to.
The next line contains the original array.
$1\leq n\leq 1000$, $-10^9\leq A[i], P\leq 10^9$。
For each test case, output one integer which means the most beautiful interval‘s beauty after your change.
2
3 5
1 -1 2
3 -2
1 -1 2
8
2
数据才1000直接暴力。。
1 #include<algorithm> 2 #include<iostream> 3 #include<cstdlib> 4 #include<cstring> 5 #include<cstdio> 6 #include<vector> 7 #include<map> 8 using std::max; 9 using std::cin; 10 using std::cout; 11 using std::endl; 12 using std::find; 13 using std::sort; 14 using std::map; 15 using std::pair; 16 using std::vector; 17 using std::multimap; 18 #define pb(e) push_back(e) 19 #define sz(c) (int)(c).size() 20 #define mp(a, b) make_pair(a, b) 21 #define all(c) (c).begin(), (c).end() 22 #define iter(c) decltype((c).begin()) 23 #define cls(arr,val) memset(arr,val,sizeof(arr)) 24 #define cpresent(c, e) (find(all(c), (e)) != (c).end()) 25 #define rep(i, n) for (int i = 1; i <= (int)(n); i++) 26 #define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i) 27 const int N = 1010; 28 const int INF = 0x7fffffff; 29 typedef long long ll; 30 ll p, ans, arr[N], sum[N]; 31 int main() { 32 #ifdef LOCAL 33 freopen("in.txt", "r", stdin); 34 freopen("out.txt", "w+", stdout); 35 #endif 36 int t, n; 37 scanf("%d", &t); 38 while (t--) { 39 ans = -INF; 40 scanf("%d %lld", &n, &p); 41 rep(i, n) scanf("%lld", &arr[i]); 42 rep(i, n) { 43 ll tmp = arr[i]; arr[i] = p; 44 rep(j, n) { 45 sum[j] = sum[j - 1] > 0 ? sum[j - 1] + arr[j] : arr[j]; 46 ans = max(ans, sum[j]); 47 } 48 arr[i] = tmp; 49 } 50 printf("%lld\n", ans); 51 } 52 return 0; 53 }
原文：http://www.cnblogs.com/GadyPu/p/4641675.html
