hdu 1003
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 -1 5 4 -7 7 0 6 -1 1 -6 7 -5
Sample Output
Case 1: 14 1 4 Case 2: 7 1 6
题意:
给你n个数a[1],a[2],a[3]...a[n],要你求出最大连续子段和,而且输出这个和,和这个子段的起始位置和终止位置,假设这个子段有多种可能,输出第一种结果就可以。
思路:
前段时间做个类似的题目,用的是暴力求解,用的两重循环。可是这个题目要求是n<100000,用两重循环肯定会超时,因此我们必须换一种思路,必须降低循环次数,当输入的时候,累计求和,用pos记录位置,star记录起点。end记录终点,假设sum小于0.就不必往以下累加了。
代码:
#include<cstdio> using namespace std; #define INF 0x7fffffff int const maxn=100000+10; int a[maxn]; int main() { int kcase=1; int T,n,max,star,end,pos,sum;//max为最大字段和。star为枚举起始位置,end为枚举终止位置。sum为字段和。pos用来更新枚举起始位置 scanf("%d",&T); while(T--) { scanf("%d",&n); max=-INF; pos=sum=0; for(int i=0;i<n;i++) { scanf("%d",&a[i]); sum+=a[i]; if(sum>max) { max=sum; star=pos; end=i; } if(sum<0) { sum=0; pos=i+1; } } printf("Case %d:\n",kcase++); printf("%d %d %d\n",max,star+1,end+1);//数组下标是从0開始的,而数的位置是从1開始的 if(T) printf("\n"); } return 0; }