Satisfactory Pairs HackerRank - pairs-again——預先處理＋不定長數組應用__初級數論

Think:
1題意：輸入n, 求(a, b)滿足a < b 且存在x, y使得x*a + y*b = n的二元組個數，要求n, a, b, x, y皆為正整數
2方法：先預先處理小於n的數的約數，然後暴力試探＋剪枝，試探符合題意的二元組個數，時間複雜度 1/x 從1到n積分，為nln(n)

vjudge題目連結

建議參考部落格１
建議參考部落格２

以下為Accepted代碼

#include <bits/stdc++.h>using namespace std;const int N = 3e5 + 4;vector <int> y_b[N];vector <int> :: iterator b;int book[N];int main(){    int n, i, j, a, x, t, ans;    scanf("%d", &n);    for(i = 1; i <= n; i++){        for(j = 1; j*j <= i; j++){            if(i%j == 0){                y_b[i].push_back(j);                if(i*i != j){                    t = i / j;                    y_b[i].push_back(t);                }            }        }        sort(y_b[i].begin(), y_b[i].end(), greater<int>());    }    ans = 0;    for(a = 1; a < n; a++){        for(x = 1; x*a < n; x++){            t = n - x*a;            for(b = y_b[t].begin(); b != y_b[t].end(); b++){                if((*b) <= a)                    break;                if(book[*b] != a){                    ans++;                    book[*b] = a;                }            }        }    }    printf("%d\n", ans);    return 0;}

以下為建議參考代碼

#include <bits/stdc++.h>using namespace std;const int N = 3e5 + 4;vector <int> y_b[N];/*記錄當前元素的約數*/vector <int> :: iterator b;int book[N];/*實數型全域變數初始化預設為0*/bool cmp(int a, int b){    return a > b;}int main(){    int n, i, j, a, x, t, ans;    scanf("%d", &n);    for(i = 1; i <= n; i++){        for(j = 1; j*j <= i; j++){            if(i%j == 0){                y_b[i].push_back(j);                if(i*i != j){                    t = i / j;                    y_b[i].push_back(t);                }            }        }        sort(y_b[i].begin(), y_b[i].end(), greater<int>());    }    ans = 0;    for(a = 1; a < n; a++){        for(x = 1; x*a < n; x++){            t = n - x*a;            for(b = y_b[t].begin(); b != y_b[t].end(); b++){                if((*b) <= a)/*題意：a < b*/                    break;                if(book[*b] != a){/*標記判重*/                    ans++;                    book[*b] = a;                }            }        }    }    printf("%d\n", ans);    return 0;}