标签：
1 /*
2 题意：求最大连续子序列和及两个端点
3 累积遍历算法 O(n)：依照sum<0将序列分块，最值在某一块上产生。dp也是同样的思路：dp[i] = max (dp[i-1] + a[i], a[i]) 其实是一样的
4 1003就这么难？？
5 详细解释
6 */
7 /************************************************
8 * Author :Running_Time
9 * Created Time :2015-8-10 17:44:42
10 * File Name :O(n).cpp
11 ************************************************/
12
13 #include <cstdio>
14 #include <algorithm>
15 #include <iostream>
16 #include <sstream>
17 #include <cstring>
18 #include <cmath>
19 #include <string>
20 #include <vector>
21 #include <queue>
22 #include <deque>
23 #include <stack>
24 #include <list>
25 #include <map>
26 #include <set>
27 #include <bitset>
28 #include <cstdlib>
29 #include <ctime>
30 using namespace std;
31
32 #define lson l, mid, rt << 1
33 #define rson mid + 1, r, rt << 1 | 1
34 typedef long long ll;
35 const int MAXN = 1e6 + 10;
36 const int INF = 0x3f3f3f3f;
37 const int MOD = 1e9 + 7;
38 int a[MAXN];
39
40 int main(void) { //HDOJ 1003 Max Sum
41 int T, cas = 0; scanf ("%d", &T);
42 while (T--) {
43 int n; scanf ("%d", &n);
44 for (int i=1; i<=n; ++i) {
45 scanf ("%d", &a[i]);
46 }
47 int sum = a[1], mx = a[1];
48 int l = 1, r = 1;
49 int ll = 1, rr = 1;
50 for (int i=2; i<=n; ++i) {
51 if (sum <= 0) {
52 sum = a[i]; l = i;
53 }
54 else sum += a[i];
55 if (mx < sum) {
56 mx = sum; ll = l, rr = i;
57 }
58 }
59
60 printf ("Case %d:\n%d %d %d\n", ++cas, mx, ll, rr);
61 if (T) puts ("");
62 }
63
64 return 0;
65 }
1 /************************************************ 2 * Author :Running_Time 3 * Created Time :2015-8-10 18:24:00 4 * File Name :On_2.cpp 5 ************************************************/ 6 7 #include <cstdio> 8 #include <algorithm> 9 #include <iostream> 10 #include <sstream> 11 #include <cstring> 12 #include <cmath> 13 #include <string> 14 #include <vector> 15 #include <queue> 16 #include <deque> 17 #include <stack> 18 #include <list> 19 #include <map> 20 #include <set> 21 #include <bitset> 22 #include <cstdlib> 23 #include <ctime> 24 using namespace std; 25 26 #define lson l, mid, rt << 1 27 #define rson mid + 1, r, rt << 1 | 1 28 typedef long long ll; 29 const int MAXN = 1e6 + 10; 30 const int INF = 0x3f3f3f3f; 31 32 int a[MAXN]; 33 34 int main(void) { 35 int T, cas = 0; scanf ("%d", &T); 36 while (T--) { 37 int n; scanf ("%d", &n); 38 for (int i=1; i<=n; ++i) { 39 scanf ("%d", &a[i]); 40 } 41 int sum = 0, mx = -INF, mn = 0; 42 int l = 0, ll = 0, rr = 1; 43 for (int i=1; i<=n; ++i) { 44 sum += a[i]; 45 if (sum - mn > mx) { 46 mx = sum - mn; 47 ll = l; rr = i; 48 } 49 if (mn > sum) { 50 mn = sum; l = i; 51 } 52 } 53 printf ("Case %d:\n%d %d %d\n", ++cas, mx, ll+1, rr); 54 if (T) puts (""); 55 } 56 57 return 0; 58 }
标签：
原文地址：http://www.cnblogs.com/Running-Time/p/4719011.html
