Day8 - G - Bound Found ZOJ - 1964

Signals of most probably extra-terrestrial origin have been received and digitalized by The Aeronautic and Space Administration (that must be going through a defiant phase: "But I want to use feet, not meters!"). Each signal seems to come in two parts: a sequence of n integer values and a non-negative integer t. We'll not go into details, but researchers found out that a signal encodes two integer values. These can be found as the lower and upper bound of a subrange of the sequence whose absolute value of its sum is closest to t.

You are given the sequence of n integers and the non-negative target t. You are to find a non-empty range of the sequence (i.e. a continuous subsequence) and output its lower index l and its upper index u. The absolute value of the sum of the values of the sequence from the l-th to the u-th element (inclusive) must be at least as close to t as the absolute value of the sum of any other non-empty range.

Input

The input file contains several test cases. Each test case starts with two numbers n and k. Input is terminated by n = k = 0. Otherwise, 1 <= n <= 100000 and there follow n integers with absolute values <= 10000 which constitute the sequence. Then follow k queries for this sequence. Each query is a target t with 0 <= t <= 1000000000.

Output

For each query output 3 numbers on a line: some closest absolute sum and the lower and upper indices of some range where this absolute sum is achieved. Possible indices start with 1 and go up to n.

Sample Input

5 1
-10 -5 0 5 10
3
10 2
-9 8 -7 6 -5 4 -3 2 -1 0
5 11
15 2
-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
15 100
0 0

Sample Output

5 4 4
5 2 8
9 1 1
15 1 15
15 1 15

思路：维护前缀和，排序后维护双指针取值，细节直接看代码

using namespace std;
#define lowbit(x) ((x)&(-x))
typedef long long LL;
typedef pair<LL, LL> PLL;
 
const int maxm = 1e5+5;

int n, k;

LL C[maxm];

void add(int x, int val) {
    for(; x <= n; x += lowbit(x))
        C[x] += val;
}

LL getsum(int x) {
    LL ret = 0;
    for(; x; x -= lowbit(x))
        ret += C[x];
    return ret;
}

struct Node {
    LL val;
    int id;
    bool operator<(const Node &a) const {
        return val < a.val;
    }
} Nodes[maxm];

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    int t, val;
    while(cin >> n >> k && n+k) {
        memset(C, 0, sizeof(C));
        for(int i = 1; i <= n; ++i) {
            cin >> val;
            add(i, val);
            Nodes[i] = {getsum(i), i};
        }
        Nodes[0].id = 0, Nodes[0].val = 0;
        sort(Nodes,Nodes+1+n);
        while(k--) {
            int l = 0, r = 1, ansL, ansR;
            LL ans, t, diff = 0x3f3f3f3f;
            cin >> t;
            while(l <= n && r <= n) {
                LL val = Nodes[r].val - Nodes[l].val;
                if(abs(val - t) < diff) {
                    ans = val;
                    diff = abs(val-t);
                    ansL = min(Nodes[l].id+1,Nodes[r].id+1);
                    ansR = max(Nodes[l].id, Nodes[r].id);
                }
                if(val > t)
                    l++;
                else if(val < t)
                    r++;
                else break;
                if(l == r) r++;
            }
            cout << ans << " " << ansL << " " << ansR << "\n";
        }
        
    }
    
    return 0;
}

View Code