BestCoder Round #39 (HDU5210,5211,5212)
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题目传送:HDU - 5210
HDU - 5210 - Delete
思路:简单题
AC代码:
#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cmath>#include <queue>#include <stack>#include <vector>#include <map>#include <set>#include <deque>#include <cctype>#define LL long long#define INF 0x7fffffffusing namespace std;int n, k; int a[105];int main() { while(scanf("%d", &n) != EOF) { set<int> se; for(int i = 0; i < n; i++) { scanf("%d", &a[i]); se.insert(a[i]); } scanf("%d", &k); int ma = se.size(); if(k <= n - ma) { printf("%d\n", ma); } else { printf("%d\n", n - k); } } return 0;}
题目传送:HDU - 5211
HDU - 5211 - Mutiple
思路:直接暴力可以过,时间复杂度直逼O(n^2),运行出来1000+ms,不过有更好的方法,先初始化每个数的最左位置为最右边(这里即n+1),再从最右往最左遍历,然后找是当前数的倍数的数中的最左位置,找完后再更新当前位置的数的最左位置,时间复杂度O(n*logn),运行时间15ms;
AC代码(暴力):
#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cmath>#include <queue>#include <stack>#include <vector>#include <map>#include <set>#include <deque>#include <cctype>#define LL long long#define INF 0x7fffffffusing namespace std;int n;int a[10005];int main() { while(scanf("%d", &n) != EOF) { for(int i = 1; i <= n; i++) { scanf("%d", &a[i]); } int ans = 0; for(int i = 1; i <= n; i++) { for(int j = i + 1; j <= n; j++) { if(a[j] % a[i] == 0) { ans += j; //printf("%d\n", ans); break; } } } printf("%d\n", ans); } return 0;}
AC代码(带优化):
#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cmath>#include <queue>#include <stack>#include <vector>#include <map>#include <set>#include <deque>#include <cctype>#define LL long long#define INF 0x7fffffffusing namespace std;int f[10005];int a[10005]; int n;int main() {while(scanf("%d", &n) != EOF) {for(int i = 1; i <= n; i ++) {scanf("%d", &a[i]);}for(int i = 1; i <= 10000; i ++) {f[i] = n + 1;//初始化每个数出现的位置为最右 }int ans = 0;for(int i = n; i >= 1; i --) {//从后往前找,并更新每个数出现的最左位置 int tmp = n + 1;for(int j = a[i]; j <= 10000; j += a[i]) {//找每个含有a[i]因子的数中的最小位置 tmp = min(tmp, f[j]);}f[a[i]] = i;//更新每个数出现的最左位置 if(tmp > n) tmp = 0;ans += tmp;}printf("%d\n", ans);}return 0;}
题目传送:Code
HDU - 5212 - Code
题目可以转化为,先求每个以i为最大公约数的两个数的对数num,然后在求(i*(i - 1))*num 的和。
然后只要快速求出以i为最大公约数的两个数的对数num即可(2 =< i <= n,因为1的时候i-1等于0了)。
可以先在n*logn的时间内求出以i为公约数的两个数的对数f[i]。然后从大往小,依次减去f[2*i],f[3*i],……,来求出以i为最大公约数的两个数的对数f[i](因为以2*i为最大公约数的两个数的对数肯定会在以i为约数的两个数的对对数中)。这样就ok了。
AC代码:
#include <map>#include <set>#include <list>#include <cmath>#include <deque>#include <queue>#include <stack>#include <bitset>#include <cctype>#include <cstdio>#include <string>#include <vector>#include <complex>#include <cstdlib>#include <cstring>#include <fstream>#include <sstream>#include <utility>#include <iostream>#include <algorithm>#include <functional>#define LL long long#define INF 0x7fffffffusing namespace std;#define MOD 10007;int n;int a[10005];//输入的数int sum[10005];//sum[i]用于统计输入的数据中i出现的次数int f[10005];//最终得出的f[i]是以i为最大公约数的两个数的对数int max_num;//输入中出现的最大数int main() {while(scanf("%d", &n) != EOF) {max_num = 0;memset(sum, 0, sizeof(sum));for(int i = 0; i < n; i ++) {scanf("%d", &a[i]);sum[a[i]] ++;max_num = max(max_num, a[i]);}memset(f, 0, sizeof(f));for(int i = 2; i <= max_num; i ++) {for(int j = i; j <= max_num; j += i) {f[i] += sum[j];}f[i] = f[i] * f[i];}int ans = 0;for(int i = max_num; i >= 2; i --) {for(int j = i + i; j <= max_num; j += i) {f[i] -= f[j];}ans = (ans + (LL) i * (i - 1) * f[i]) % MOD;}printf("%d\n", ans);}return 0;}
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