Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others) Total Submission(s): 23856 Accepted Submission(s): 6048
Problem Description
Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X.
Input
There are many cases. Every data case is described as followed: In the first line there are three integers L, N, M, in the second line there are L integers represent the sequence A, in the third line there are N integers represent the sequences B, in the forth line there are M integers represent the sequence C. In the fifth line there is an integer S represents there are S integers X to be calculated. 1<=L, N, M<=500, 1<=S<=1000. all the integers are 32-integers.
Output
For each case, firstly you have to print the case number as the form "Case d:", then for the S queries, you calculate if the formula can be satisfied or not. If satisfied, you print "YES", otherwise print "NO".
Sample Input
3 3 3 1 2 3 1 2 3 1 2 3 3 1 4 10
Sample Output
Case 1: NO YES NO
#include#include #include #define MAX_N 500050using namespace std;const double ESP = 1e-5;const int INF = 1e8;int L[MAX_N], N[MAX_N], M[MAX_N], LN[MAX_N];int M1[MAX_N];int l, n, m, cnt;bool lower_bound(int x) { int lp = -1, rp = cnt - 1; while (rp - lp > 1) { int mid = (lp + rp)/2; if (LN[mid] >= x) rp = mid; else lp = mid; } //printf("%d %d\n", x, LN[rp]); return LN[rp] == x;}int main() { int t, p; int cut = 0; while (scanf("%d%d%d", &l, &n, &m) != EOF) { printf("Case %d:\n", ++cut); for (int i = 0; i < l; i++) scanf("%d", &L[i]); for (int i = 0; i < n; i++) scanf("%d", &N[i]); for (int i = 0; i < m; i++) scanf("%d", &M1[i]); //利用等式L[i]+N[i] = x - M[i]降低运行时间 cnt = 0; for (int i = 0; i < l; i++) { for (int j = 0; j < n; j++) { LN[cnt++] = L[i] + N[j]; } } sort(LN, LN + cnt); //printf("%d\n", LN[0]); scanf("%d", &t); while (t--) { bool flag = false; scanf("%d", &p); //用p减去M中的每一个元素 for (int i = 0; i < m; i++) { M[i] = p - M1[i]; } sort(M, M + m); for (int i = 0; i < m; i++) { if (lower_bound(M[i])) { flag = true; break; } } if (flag) printf("YES\n"); else printf("NO\n"); } } return 0;}