PAT甲级 1007
来源:互联网 发布:淘宝怎么开通蚂蚁花呗 编辑:程序博客网 时间:2024/06/02 03:15
Maximum Subsequence Sum
Given a sequence of K integers { N1, N2, …, NK }. A continuous subsequence is defined to be { Ni, Ni+1, …, Nj } where 1 <= i <= j <= K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.
Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.
Input Specification:
Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K (<= 10000). The second line contains K numbers, separated by a space.
Output Specification:
For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.
Sample Input:
10 -10 1 2 3 4 -5 -23 3 7 -21
Sample Output:
10 1 4
判断条件后(逻辑推理),对tempRecord,retRecord进行更新。注意题意和输出。
#include<iostream>#include<stdio.h>using namespace std;const int maxn=10000+1;int input[maxn];typedef struct node{ int i,j; int sum; node(int ti,int tj,int tsum){ i=ti; j=tj; sum=tsum; }}record;int main(){ //freopen("./in","r",stdin); int k; scanf("%d",&k); int i; for(i=0;i<k;i++){ scanf("%d",input+i); } record tempRecord(0,0,0),retRecord(0,0,input[0]); for(i=0;i<k;i++){ tempRecord.sum+=input[i]; if(input[i]<0 && tempRecord.sum<0){ if(retRecord.sum<tempRecord.sum){ retRecord.sum=tempRecord.sum; retRecord.i=retRecord.j=i; } tempRecord.sum=0; tempRecord.i=i+1; }else if(input[i]>=0 && tempRecord.sum>retRecord.sum){ retRecord.sum=tempRecord.sum; retRecord.i=tempRecord.i; retRecord.j=tempRecord.j=i; } } if(retRecord.sum>=0){ printf("%d %d %d",retRecord.sum,input[retRecord.i],input[retRecord.j]); }else{ printf("%d %d %d",0,input[0],input[k-1]); }}
- PAT(甲级)1007
- PAT甲级1007
- PAT 甲级1007
- PAT-甲级-1007
- PAT甲级1007
- PAT甲级1007
- PAT甲级 1007
- 浙大PAT甲级-1007
- 【PAT】甲级1007
- PAT甲级 1007
- PAT 甲级
- PAT 1007(甲级)Maximum Subsequence Sum
- PAT甲级 A1025.PAT RANKING
- PAT 甲级 1025 PAT Ranking
- PAT(甲级)1003
- PAT(甲级)1004
- PAT(甲级)1005
- PAT(甲级)1006
- web里面html,css,js基本知识
- 软件工程师具备六大基本素质
- java获取URL响应头的日期时间
- jQuery
- 类和对象运行时在内存里是怎么样的?各种变量、方法在运行时是怎么交互的?
- PAT甲级 1007
- redis介绍及安装及redis 集群
- 刷清橙OJ--A1077.坐标排序
- jQuery
- 访问国外网站跳转到CJB的解决办法
- Linux基础(二)
- 用回调函数实现冒泡排序
- 文件路径
- 2017.6.19c++