CodeForces 484B Maximum Value
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <vector> 6 #define sc(x) scanf("%d", &x) 7 #define sc2(x,y) scanf("%d%d", &x, &y) 8 #define pf(x) printf("%d\n",x) 9 #define PF(x) printf("%d ",x) 10 #define FOR(i,b,e) for(int i=b;i<=e;i++) 11 #define FOR1(i,b,e) for(int i=b;i>=e;i--) 12 #define CL(x,y) memset(x,y,sizeof(x)) 13 using namespace std; 14 const int MAX = 1000000; 15 vector <int> v; 16 int n; 17 long long used[MAX]; 18 int main() 19 { 20 int tmp, ans; 21 while(~sc(n)) 22 { 23 ans = 0; 24 CL(used, 0); 25 FOR(i,0,n-1) 26 { 27 sc(tmp); 28 used[tmp] = 1; 29 v.push_back(tmp); 30 } 31 sort(v.begin(), v.end()); 32 FOR(i,0,n-2) 33 { 34 if(v[i] == 0 || v[i] == v[i+1]) 35 continue; 36 FOR(j,i+1,n-1) 37 { 38 if(used[v[j]]) 39 ans = ans > (v[j] % v[i]) ? ans : (v[j] % v[i]); 40 used[v[j]] = 0; 41 } 42 } 43 pf(ans); 44 } 45 return 0; 46 }
以上代码超时，不可以
方法二：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <vector> 6 #define sc(x) scanf("%d", &x) 7 #define sc2(x,y) scanf("%d%d", &x, &y) 8 #define pf(x) printf("%d\n",x) 9 #define PF(x) printf("%d ",x) 10 #define FOR(i,b,e) for(int i=b;i<=e;i++) 11 #define FOR1(i,b,e) for(int i=b;i>=e;i--) 12 #define CL(x,y) memset(x,y,sizeof(x)) 13 using namespace std; 14 const int MAX = 1000000; 15 vector <int> v; 16 int n, used[MAX+5]; 17 int main() 18 { 19 int tmp, ans; 20 while(~sc(n)) 21 { 22 ans = 0; 23 CL(used,0); 24 FOR(i,0,n-1) 25 { 26 sc(tmp); 27 if(!used[tmp])//这里进行了取出重复的 28 { 29 v.push_back(tmp); 30 used[tmp] = true; 31 } 32 } 33 sort(v.begin(), v.end()); 34 n = v.size(); 35 FOR(i,0,n-1) 36 { 37 if(v[i] == 0) continue; 38 int pos = 0, k = 2, num = v[i]; 39 while(pos<n-1) 40 { 41 pos = lower_bound(v.begin(), v.end(), num * k++) - v.begin() - 1;//最多就是插在尾部(n的位置) 42 tmp = v[pos] % num; 43 ans = ans > tmp ? ans : tmp; 44 if(pos == n-1) 45 { 46 break; 47 } 48 } 49 } 50 pf(ans); 51 } 52 return 0; 53 }
这里减少了动态数组中存在的数，减少了不必要的判断
方法三：
1 /********************************/ 2 /*Problem: */ 3 /*User: shinelin */ 4 /*Memory: 1000K */ 5 /*Time: 1000MS */ 6 /*Language: C++ */ 7 /********************************/ 8 #include <cstdio> 9 #include <iostream> 10 #include <cstdlib> 11 #include <ctime> 12 #include <cctype> 13 #include <cstring> 14 #include <string> 15 #include <list> 16 #include <map> 17 #include <queue> 18 #include <deque> 19 #include <stack> 20 #include <vector> 21 #include <set> 22 #include <algorithm> 23 #include <cmath> 24 using namespace std; 25 26 #define INF 0x7fffffff 27 #define LL long long 28 29 vector<int> a; 30 int N; 31 32 bool BigMod(int d) 33 { 34 int k = lower_bound(a.begin(), a.end(), d) - a.begin(); 35 for (int i = k; i < N; i ++) 36 { 37 if(a[i] == a[i-1]) continue; 38 int j = 2 * a[i]; 39 while (j <= a[N-1]) 40 { 41 int cur = lower_bound(a.begin(), a.end(), j) - a.begin(); 42 if(cur > 0 && a[cur-1] % a[i] > d) 43 { 44 return true; 45 } 46 j += a[i]; 47 } 48 if(a[N-1] % a[i] > d) 49 return true; 50 } 51 return false; 52 } 53 int main() 54 { 55 int x, Min, Max, Mid; 56 scanf("%d", &N); 57 for (int i = 0; i < N; i ++) 58 { 59 scanf("%d", &x); 60 a.push_back(x); 61 } 62 sort(a.begin(), a.end()); 63 Min = 0; 64 Max = a[N-1]; 65 while(Min < Max) 66 { 67 Mid = (Max + Min) >> 1; 68 if(BigMod(Mid)) 69 { 70 Min = Mid + 1; 71 } 72 else 73 { 74 Max = Mid; 75 } 76 } 77 // Mid = (Max + Min) >> 1; 78 printf("%d\n", Min); 79 return 0; 80 }
必须真正有二分的思想解决（没看懂呢）
