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  • PAT1007. Maximum Subsequence Sum

    https://www.patest.cn/contests/pat-a-practise/1007

    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

    题目输出要看仔细,处理0是个坑
    贪心就好,强行dp很别扭
    // 去吧!皮卡丘! 把AC带回来!
    //      へ     /|
    //   /\7    ∠_/
    //   / │   / /
    //  │ Z _,< /   /`ヽ
    //  │     ヽ   /  〉
    //  Y     `  /  /
    //  イ● 、 ●  ⊂⊃〈  /
    //  ()  へ    | \〈
    //   >ー 、_  ィ  │ //
    //   / へ   / ノ<| \\
    //   ヽ_ノ  (_/  │//
    //    7       |/
    //    >―r ̄ ̄`ー―_
    //**************************************
    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include <bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    #define inf 2147483647
    const ll INF = 0x3f3f3f3f3f3f3f3fll;
    #define ri register int
    template <class T> inline T min(T a, T b, T c) { return min(min(a, b), c); }
    template <class T> inline T max(T a, T b, T c) { return max(max(a, b), c); }
    template <class T> inline T min(T a, T b, T c, T d) {
      return min(min(a, b), min(c, d));
    }
    template <class T> inline T max(T a, T b, T c, T d) {
      return max(max(a, b), max(c, d));
    }
    #define scanf1(x) scanf("%d", &x)
    #define scanf2(x, y) scanf("%d%d", &x, &y)
    #define scanf3(x, y, z) scanf("%d%d%d", &x, &y, &z)
    #define scanf4(x, y, z, X) scanf("%d%d%d%d", &x, &y, &z, &X)
    #define pi acos(-1)
    #define me(x, y) memset(x, y, sizeof(x));
    #define For(i, a, b) for (int i = a; i <= b; i++)
    #define FFor(i, a, b) for (int i = a; i >= b; i--)
    #define bug printf("***********
    ");
    #define mp make_pair
    #define pb push_back
    const int maxn = 5e5 + 10;
    // name*******************************
    int n;
    int sum = 0;
    int x;
    int h, t;
    int ans = 0;
    int a[10005];
    
    // function******************************
    
    //***************************************
    int main() {
      // ios::sync_with_stdio(0);
      // cin.tie(0);
      // freopen("test.txt", "r", stdin);
      //  freopen("outout.txt","w",stdout);
      cin >> n;
      int h1 = 1;
      h = 1, t = n;
      bool flag1 = true, flag2 = true;
      For(i, 1, n) {
        cin >> a[i];
        if (a[i] > 0)
          flag1 = false;
        if (a[i] == 0 && flag2)
          flag2 = false;
        sum += a[i];
        if (sum < 0) {
          sum = 0;
          h1 = i + 1;
        }
        if (sum > ans) {
          h = h1;
          t = i;
          ans = sum;
        }
      }
      if (flag1) {
        if (flag2)
          cout << 0 << " " << a[1] << " " << a[n];
        else
          cout << "0 0 0";
      } else
        cout << ans << " " << a[h] << " " << a[t];
    
      return 0;
    }
    未优化代码
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  • 原文地址:https://www.cnblogs.com/planche/p/8570003.html
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