Sdut 1351 Max Sum

Max Sum Time limit:2000ms Memory limit:32768k have questions? Dot here ^_^ Title 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. Enter the first line of the input cont Ains 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 is b etween-1000 and 1000). Output for each test case, you should output both lines. The first line was "Case #:", # means the number of the the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end posi tion of the sub-sequence. If there is more than one result, output the first one. Output a blank line between cases. Example input

25 6-1 5 4-77 0 6-1 1-6 7-5

Sample output

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

Tip hdoj1003 Note: The background test data volume is large, please use the C language scanf input. Avoid timeouts!!!! SOURCE IGNATIUS.L

#include <iostream> #include <algorithm> #include <stdio.h> #include <string.h> #include <    Stdlib.h> #include <queue>using namespace Std;int a[100100],b[100100];int main () {int T;    int k = 0;    scanf ("%d", &t);        while (t--) {int n;        scanf ("%d", &n);        for (int i=0;i<n;i++) {scanf ("%d", &a[i]);        } B[0] = a[0];            for (int i=1;i<n;i++) {if (b[i-1]<0) {b[i] = A[i];            } else {B[i] = A[i] + b[i-1];        }} int xx,yy;        int maxx =-999999;                for (int i=0;i<n;i++) {if (Maxx<b[i]) {Maxx = B[i];            yy = i+1;        }} int pmax = 0;            for (int i=yy-1;i>=0;i--) {Pmax + = A[i];            printf ("Pmax =%d Maxx =%d\n", Pmax,maxx);        if (Pmax = = Maxx)    {xx = i+1;            Break        }} printf ("Case%d:\n", ++k);        printf ("%d%d%d\n", maxx,xx,yy);        if (T) {printf ("\ n");  }} return 0;}

Sdut 1351 Max Sum