題目連結
題目大意：
給一set S, 有多個instinct integers, 求最大的d使得a+b+c = d
解題思路
將算式變作a+b = c-d。開一陣列x記錄所有a+b，用qsort排列x的index。再算出每一種可能的c-d，用binary search搜尋x，若value相同且c, d不等於a, b，則紀錄d，最後找出最大d，無解則輸出no solution。
Code
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| #include <cstdio>#include <stdlib.h>#include <algorithm>using namespace std;int elements[1000];struct a_plus_b{ int a; int b; int sum;};struct a_plus_b x[1000000];int search_comp(int d, int c, int x_index, int x_index_tb[]) { int i,lower_bound=0, upper_bound=x_index-1, mid; struct a_plus_b value; int y = d-c; while(lower_bound <= upper_bound) { mid = (lower_bound+upper_bound)/2; value = x[x_index_tb[mid]]; if(value.sum > y) upper_bound = mid-1; else if(value.sum < y) lower_bound = mid+1; else { for(i = mid; i <= upper_bound; i++) { if(value.sum != y) break; if(d!=value.a && d!=value.b && c!=value.a && c!=value.b) return 1; } for(i = mid-1; i >= lower_bound; i--) { if(value.sum != y) break; if(d!=value.a && d!=value.b && c!=value.a && c!=value.b) return 1; } return 0; } } return 0;}int comp_func(const void *a, const void *b){ int c = x[*(int*)a].sum; int d = x[*(int*)b].sum; if (c>d) return 1; else if (c< d) return -1; else return 0;}int main(){ int i, j, element_num; while(scanf("%d",&element_num)==1 && element_num){ int x_index = 0; int x_index_tb[element_num*(element_num-1)/2]; for (i=0; i< element_num; i++) scanf("%d", &elements[i]); for (i=0; i< element_num; i++){ for (j=i+1; j< element_num; j++){ x[x_index].sum = elements[i]+elements[j]; x[x_index].a = elements[i]; x[x_index].b = elements[j]; x_index_tb[x_index] = x_index; x_index++; } } qsort(x_index_tb, x_index, sizeof(int), comp_func); int y1bs, max=0, flag=0; for (i=element_num-1; i>=0; i--){ for (j=0; j< element_num; j++){ if (i==j) continue; y1bs = search_comp(elements[i], elements[j], x_index, x_index_tb); if(y1bs == 1) { if (flag==0){ max = elements[i]; flag = 1; } else if (elements[i]>max) max = elements[i]; } } } if (flag==1) printf("%d\n",max); else printf("no solution\n"); } return 0;} |
