star 比較 esc 思路 ++ n) scan input script

題目：

Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

Input The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

Output For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

Sample Input 2 5 6 -15 4 -7 7 0 6 -1 1 -6 7 -5

Sample Output Case 1: 14 1 4 Case 2: 7 1 6 題意描述：
輸入一串數字為N N(1<=N<=100000)和N個範圍-1000到1000的整型數 計算並輸出該串數字的最大子串（連續）和以及該子串的起始位置和終止位置 解題思路：
該題還是比較考察思維的，屬於簡單的DP問題。遍歷每個元素，將其累加到sum上，並每次判斷sum是否大於max，若大於max則更新sum後，將始末位置也更新一下；若sum<0,表示該數是一個很大的負數需要重新開始累加sum，尋找下一個可能存在的最大子串和。 代碼實現：

 1 #include<stdio.h>
 2 int main()
 3 {
 4     int T,t,n,a[100010],i,sum,s,f,c=0,max;
 5     scanf("%d",&T);
 6     while(T--)
 7     {
 8         scanf("%d",&n);
 9         for(i=1;i<=n;i++)
10             scanf("%d",&a[i]);
11         sum=0;
12         max=-99999999;//註意將max賦值為一個較小的賦值 
13         t=1;
14         for(i=1;i<=n;i++){
15             sum += a[i];
16             if(sum > max){
17                 max=sum;
18                 s=t;
19                 f=i;
20             }
21             if(sum < 0){
22                 sum=0;
23                 t=i+1;
24             }
25         }
26         printf("Case %d:\n%d %d %d\n",++c,max,s,f);
27         if(T != 0)
28         printf("\n");
29     }
30     return 0;
31 } 

易錯分析：

1、註意格式問題

2、註意max的初始化

HDU 1003 Max Sum