hdu1003 Max sum

题目

 

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 -1 5 4 -7 7 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1: 14 1 4 Case 2: 7 1 6

 

分析

动规最大子段和问题，比较模板

代码

/**********************
User:Mandy.H.Y
Language:c++
Problem:hdu max sum
Algorithm:
Scores:
**********************/
//hdu与poj都不能用万能头文件 
//#include<bits/stdc++.h>
#include<cstdio> 
#include<iomanip> 
#include<algorithm> 
#include<cmath> 

using namespace std;

const int maxn = 1e5 + 5;

int t,n,ans,ansl,ansr;
int a[maxn];
int l[maxn],r[maxn],sum[maxn];

template<class T>inline void read(T &x){
    x = 0;bool flag = 0;char ch = getchar();
    while(!isdigit(ch)) flag |= ch == '-',ch = getchar();
    while(isdigit(ch)) x = (x << 3) + (x << 1) + (ch ^ 48),ch = getchar();
    if(flag) x = -x;
}

template<class T>void putch(const T x){
    if(x > 9) putch(x / 10);
    putchar(x % 10 | 48);
}

template<class T>void put(const T x){
    if(x < 0) putchar('-'),putch(-x);
    else putch(x);
}

void readdata(){
    read(n);
    for(int i = 1;i <= n; ++ i) read(a[i]);
}

void work(int j){
    ans = -2e9;
    sum[0] = -1;//初始化 
    for(int i = 1;i <= n;++ i){
        l[i] = i;r[i] = i;
        if(sum[i - 1] >= 0){//要输出第一个位置，需要sum[0]初始化为负 
            sum[i] = sum[i - 1] + a[i];
            l[i] = l[i - 1];
        }else sum[i] = a[i];
        
        if(ans < sum[i]){//要输出第一个位置 
            ans = sum[i];
            ansl = l[i];
            ansr = r[i];
        }
    }
    
    printf("Case %d:\n",j);
    put(ans);putchar(' ');
    put(ansl);putchar(' ');
    put(ansr);
    puts("");//一个空格和换行符引发的血案 
    if(j != t) puts("");//要空行 
}

int main(){
    read(t);
    for(int i = 1;i <= t; ++ i){
        readdata();
        work(i);
    }
    return 0;
}