【動態規劃or雙指針】PAT 1007 Maximum Subsequence Sum

Given a sequence of K integers { N​1​​, N​2​​, ..., N​K​​ }. A continuous subsequence is defined to be { N​i​​, N​i+1​​, ..., N​j​​ } 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

方法一：動態規劃

注意，要求的是最小的i和j，所以在更新最大值的時候，只有在>才更新，如果=就不更新。而且用下面的方法求最左邊的數的話，如果有前導零，則應該把0也算在序列中。

#include<stdio.h>
#include<vector>
using namespace std;
const int maxn = 1e4+5;
const int inf = 0x3f3f3f3f;
int main(){
    int n;
    vector<int> vt;
    scanf("%d", &n);
    for(int i = 0; i < n; i++){
        int d;
        scanf("%d", &d);
        vt.push_back(d);
    }
    int dp[maxn];
    dp[0] = vt[0];
    int res = -inf, l, r;
    for(int i = 1; i < n; i++){
        if(dp[i-1] <= 0){
            dp[i] = vt[i];
        }
        else{
            dp[i] = dp[i-1]+vt[i];
        }
    }
    int index = 0;
    for(int i = 0; i < n; i++){
        if(dp[i] > res){
            res = dp[i];
            index = i;
        }
    }
    if(res < 0){
        printf("0 %d %d", vt[0], vt[n-1]);
        return 0;
    }
    printf("%d", res);
    int tindex = index;
    while(res){
        res -= vt[tindex];
        tindex--;
    }
    while(tindex >= 0 && vt[tindex] == 0) tindex--;
    printf(" %d %d", vt[min(tindex+1, n-1)], vt[index]);

    return 0;
}

方法二：雙指針

 注意，賦給sum的初值爲-1，而不能是0，因爲0也是一個“合法”的最大和，比如如果有一個序列爲：-1 0 -2 -3 -4，那麼最大和爲0，應該輸出0 0 0，而如果sum的初值是0的話，那麼if(temp > sum)裏的代碼段將不會被執行，l和r將不會更新，將輸出第一個和最後一個數。而如果把if(temp > sum)改爲if(temp >= sum)，則題目中輸出最小的i和j的要求將不能滿足。

#include<stdio.h>
using namespace std;
const int maxn = 1e4+5;
int main(){
    int num[maxn];
    int n;
    scanf("%d", &n);
    int l = 0, r = n-1, temp = 0, sum = -1, tempindex = 0;
    for(int i = 0; i < n; i++){
        scanf("%d", num+i);
        temp += num[i];
        if(temp < 0){
            temp = 0;
            tempindex = i+1;
        }
        else{
            if(temp > sum){
                sum = temp;
                r = i;
                l = tempindex;
            }
        }
    }
    if(sum < 0) sum = 0;
    printf("%d %d %d", sum, num[l], num[r]);

    return 0;
}