每次读题都是伤.....
Math
1 #include <iostream> 2 using namespace std; 3 4 int main() 5 { 6 int n, m, a; 7 long long ans, x, y; 8 cin >> n >> m >> a; 9 x = n/a; 10 y = m/a; 11 if(n%a!=0) x++; 12 if(m%a!=0) y++; 13 ans = x*y; 14 cout << ans << endl; 15 return 0; 16 }
1 #include<iostream> 2 #include<algorithm> 3 #include<string.h> 4 using namespace std; 5 6 int main() 7 { 8 int n, a, all = 0,ch1[105],ch2[105]; 9 memset(ch1,0,sizeof(ch1)); 10 memset(ch2,0,sizeof(ch2)); 11 cin >> n; 12 for(int i = 0; i < n; i++) 13 { 14 cin >> a; 15 ch1[a]++; 16 } 17 for(int i = 0; i < n; i++) 18 { 19 cin >> a; 20 ch2[a]++; 21 } 22 for(int i = 1; i <= 5; i++) 23 { 24 if((ch1[i]+ch2[i])%2 != 0) 25 { 26 cout << "-1 "; 27 return 0; 28 } 29 if((ch1[i] + ch2[i])==0) 30 continue; 31 if((ch1[i]!=ch2[i])) 32 { 33 all += abs((ch1[i] - ch2[i])/2); 34 } 35 } 36 cout << all/2 << endl; 37 return 0; 38 }
1 #include <iostream> 2 #include <algorithm> 3 using namespace std; 4 5 struct stu{ 6 int a, b; 7 }p[200010]; 8 9 int cmpa(stu x, stu y){ 10 if(x.a==y.a) 11 return x.b < y.b; 12 return x.a < y.a; 13 } 14 15 int cmpb(stu x, stu y){ 16 if(x.b==y.b) 17 return x.a < y.a; 18 return x.b < y.b; 19 } 20 21 long long solve(int a){ 22 if(a < 2) return 0; 23 return (long long)a*(a-1)/2; 24 } 25 26 int main() 27 { 28 int n; 29 cin >> n; 30 for(int i = 0; i < n; i++) 31 cin >> p[i].a >> p[i].b; 32 sort(p, p+n, cmpa); 33 int cnt = 1, t = 0, num = 1; 34 long long ans = 0; 35 for(int i = 1; i < n; i++) 36 { 37 if(p[i].a == p[t].a) 38 cnt++; 39 else 40 { 41 ans += solve(cnt); 42 t = i; 43 cnt = 1; 44 } 45 } 46 ans += solve(cnt); 47 48 sort(p, p+n, cmpb); 49 t = 0, cnt = 1; 50 for(int i = 1; i < n; i++) 51 { 52 if(p[i].a == p[i-1].a && p[i].b == p[i-1].b) 53 num++; 54 else 55 { 56 ans -= solve(num); 57 num = 1; 58 } 59 60 if(p[i].b == p[t].b) 61 cnt++; 62 else 63 { 64 ans += solve(cnt); 65 t = i; 66 cnt = 1; 67 } 68 } 69 ans -= solve(num); 70 ans += solve(cnt); 71 cout << ans << endl; 72 return 0; 73 }
Dp
1 #include <stdio.h> 2 long long d[100010], ans[100010]; 3 int main() 4 { 5 int n,k,a,b; 6 scanf("%d%d", &n, &k); 7 for(int i = 0; i < k; i++) 8 d[i] = 1; 9 ans[0] = 0; 10 for(int i = k; i < 100010; i++) 11 { 12 d[i] = (d[i-1] + d[i-k])%1000000007; 13 } 14 for(int i = 1; i < 100010; i++) 15 { 16 ans[i] = (ans[i-1] + d[i])%1000000007; 17 } 18 while(n--) 19 { 20 scanf("%d%d", &a, &b); 21 printf("%I64d ", (1000000007+ans[b]-ans[a-1])%1000000007); 22 } 23 return 0; 24 }
1 #include<iostream> 2 #include<string.h> 3 #define MAX 100010 4 using namespace std; 5 6 7 int main() 8 { 9 int n, ch[MAX],b[MAX]; 10 cin >> n; 11 memset(ch,0,sizeof(ch)); 12 memset(b,0,sizeof(ch)); 13 for(int i = 1; i <= n; i++) 14 { 15 cin >> ch[i]; 16 if(ch[i] == ch[i-1]) 17 b[i] = b[i-1];//b[i]表示上一个不同于该元素的元素的位置 18 else b[i] = i-1; 19 } 20 int ans =0; 21 for(int i= n; i > 0; i--) 22 { 23 int len = 1, j = i-1; 24 int num1 = ch[i]; 25 int num2 = MAX; 26 while(j>0) 27 { 28 if(num1 == ch[j] || num2 == ch[j]) 29 { 30 len += j-b[j]; 31 j = b[j]; 32 } 33 else if(num2 == MAX) 34 { 35 num2 = ch[j]; 36 len += j-b[j]; 37 j = b[j]; 38 } 39 else break; 40 } 41 ans = max(ans, len); 42 if(i < ans) break; 43 } 44 cout << ans << endl; 45 return 0; 46 }
Brute_forces
1 #include <iostream> 2 #include <set> 3 using namespace std; 4 5 int ch1[105], ch2[105], ch3[105]; 6 7 int main() 8 { 9 int n; 10 while(cin >> n) 11 { 12 int x=0, y=0, z=0, a; 13 for(int i = 0; i < n; i++) 14 { 15 cin >> a; 16 if(a == 0) ch3[z++] = a; 17 else if(a > 0) 18 { 19 ch2[y++] = a; 20 } 21 else ///a<0 22 { 23 ch1[x++] = a; 24 } 25 } 26 if(y == 0) 27 { 28 ch2[y++] = ch1[--x]; 29 ch2[y++] = ch1[--x]; 30 } 31 if(x % 2 == 0) 32 { 33 ch3[z++] = ch1[--x]; 34 } 35 36 cout << x; 37 for(int i = 0; i < x; i++) 38 cout << " " << ch1[i]; 39 cout << endl; 40 41 cout << y; 42 for(int i = 0; i < y; i++) 43 cout << " " << ch2[i]; 44 cout << endl; 45 46 cout << z; 47 for(int i = 0; i < z; i++) 48 cout << " " << ch3[i]; 49 cout << endl; 50 } 51 return 0; 52 }
1 #include <iostream> 2 using namespace std; 3 // 4 int ch[200010]; 5 6 int main() 7 { 8 int n, t, c; 9 cin >> n >> t >> c; 10 for(int i = 0; i < n; i++) 11 cin >> ch[i]; 12 int i = 0, cou = 0, ans = 0; 13 while(i < n) 14 { 15 while(ch[i] <= t && i < n && cou < c) 16 { 17 cou++; 18 i++; 19 } 20 if(cou == c) 21 { 22 cou--; 23 ans++; 24 } 25 else 26 { 27 i++; 28 cou = 0; 29 } 30 } 31 cout << ans <<endl; 32 return 0; 33 }
Greedy
1 /* 2 首先根据n^2的思路，我们用出现次数最多的数字来生成这个数列， 3 即用它尽量构造长度越长的连续子数列，这样一定是最优的，因为 4 这样一个数字若在中间可以同时影响其前后两个数字，举个例子: 5 234这三个数字可以得到2个ai<ai+1的组数，而若不连续，2356， 6 需要4个数才可以得到。根据这个构造思路我们还可以得到最后 7 如果有剩下的数字那必然是出现次数最多的那个，假设现在出 8 现次数最多的数字为a，a出现了cnt次，满足ai<ai+1的组数 9 为ans=n-cnt。现在再加一个数字b进去，若a==b则按照最 10 优构造思路将其插到连续的a里，这样组数不变相当于 11 ans = (n + 1) - (cnt + 1) = n - cnt，若a!=b则 12 必然可以拿出来一个a和b组成一组， 13 即(n + 1) - cnt = n - cnt + 1 = ans + 1， 14 根据这样的归纳可以证得结论的正确性 15 16 #include <cstdio> 17 #include <algorithm> 18 using namespace std; 19 int cnt[1005], n, tmp, ma; 20 21 int main() 22 { 23 scanf("%d", &n); 24 for(int i = 0; i < n; i++) 25 { 26 scanf("%d", &tmp); 27 ma = max(ma, ++ cnt[tmp]); 28 } 29 printf("%d ", n - ma); 30 } 31 */ 32 33 #include <stdio.h> 34 #include <queue> 35 #include <vector> 36 using namespace std; 37 38 int main() 39 { 40 int n, a, b; 41 priority_queue<int, vector<int>, greater<int> > q1; 42 priority_queue<int, vector<int>, greater<int> > q2; 43 scanf("%d", &n); 44 for(int i = 0; i < n; i++) 45 { 46 scanf("%d", &a); 47 q1.push(a); 48 } 49 int ans = 0; 50 while(q1.size() || q2.size()) 51 { 52 if(q1.size()) 53 { 54 a = q1.top(); 55 q1.pop(); 56 } 57 while(q1.size()) 58 { 59 b = q1.top(); 60 q1.pop(); 61 if(b > a) 62 { 63 ans++; 64 a = b; 65 } 66 else q2.push(b); 67 } 68 if(q2.size()) 69 { 70 a = q2.top(); 71 q2.pop(); 72 } 73 while(q2.size()) 74 { 75 b = q2.top(); 76 q2.pop(); 77 if(b>a) 78 { 79 ans++; 80 a = b; 81 } 82 else q1.push(b); 83 } 84 } 85 printf("%d ", ans); 86 return 0; 87 }
Constructive Algorithms
1 #include <iostream> 2 using namespace std; 3 4 long long ch[502][502];//多个数据相加 5 6 int main() 7 { 8 int n; 9 cin >> n; 10 if(n == 1) 11 {//特殊情况，其实感觉这里填什么正数都可以啊 12 cin >> n; 13 cout << "1" <<endl; 14 } 15 else 16 { 17 bool flag = false, now = true; 18 int x,y; 19 for(int i = 0; i < n; i++) 20 for(int k = 0; k < n; k++) 21 { 22 cin >> ch[i][k]; 23 if(ch[i][k] == 0) 24 { 25 x = i; 26 y = k; 27 } 28 } 29 if(x == n-y-1 || x == y) flag = true;//在对角线上 30 long long sum = 0; 31 if(!flag) 32 { 33 for(int i = 0; i < n; i++) 34 sum += ch[i][i]; 35 } 36 else 37 { 38 if(x == 0) 39 for(int i = 0; i < n; i++) 40 sum += ch[1][i]; 41 else 42 for(int i = 0; i < n; i++) 43 sum += ch[0][i]; 44 } 45 46 long long sum1 = 0; 47 for(int i = 0; i < n; i++) 48 sum1 += ch[x][i]; 49 ch[x][y] = sum - sum1; 50 //··············检查 51 52 for(int i = 0; i < n; i++) 53 { 54 sum1 = 0; 55 for(int k = 0; k < n; k++) 56 sum1 += ch[i][k]; 57 if(sum1 != sum) 58 { 59 now = false; 60 } 61 } 62 63 for(int i = 0; i < n; i++) 64 { 65 sum1 = 0; 66 for(int k = 0; k < n; k++) 67 sum1 += ch[k][i]; 68 if(sum1 != sum) 69 { 70 now = false; 71 } 72 } 73 sum1 = 0; 74 for(int i = 0; i < n; i++) 75 sum1 += ch[i][i]; 76 if(sum1 != sum) 77 { 78 now = false; 79 } 80 sum1 = 0; 81 for(int i = 0; i < n; i++) 82 sum1 += ch[i][n-i-1]; 83 if(sum1 != sum) 84 { 85 now = false; 86 } 87 if(now && ch[x][y] > 0) 88 cout << ch[x][y]; 89 else cout << "-1 "; 90 } 91 92 return 0; 93 }
implementation
1 #include <iostream>//没读懂题ac系列0.0 找到后面最小的一个交换 2 using namespace std; 3 4 int ch[3009],sh[3009][2]; 5 6 int main() 7 { 8 int n; 9 cin >> n; 10 for(int i =0; i < n; i++) 11 cin >> ch[i]; 12 int cnt = 0, mid; 13 for(int i = 0; i < n; i++) 14 { 15 int now = i; 16 for(int k = i+1; k < n; k++) 17 { 18 if(ch[now]>ch[k]) 19 { 20 now = k; 21 } 22 } 23 if(now!=i) 24 { 25 sh[cnt][0] = i; 26 sh[cnt][1] = now; 27 mid = ch[i]; 28 ch[i] = ch[now]; 29 ch[now] = mid; 30 cnt++; 31 } 32 } 33 cout << cnt <<endl; 34 for(int i = 0; i < cnt; i++) 35 cout << sh[i][0] << " " <<sh[i][1] <<endl; 36 return 0; 37 }
1 #include <iostream> 2 using namespace std; 3 4 int main() 5 { 6 long long k; 7 int n,ch[100010]; 8 cin >> n >> k; 9 for(int i = 0; i < n; i++) 10 cin >> ch[i]; 11 long long num = 0; 12 long long i; 13 for(i = 1; i <= n; i++) 14 { 15 num+=i; 16 if(num == k) 17 { 18 cout << ch[i-1]<<endl; 19 return 0; 20 } 21 else if(num > k) 22 break; 23 } 24 cout << ch[k-(num-i)-1] << endl; 25 return 0; 26 }
Graph
1 #include <iostream> 2 #include <string.h> 3 #include <vector> 4 #include <queue> 5 #define MAX 100010 6 #define INF 9223372036854775800//因为是long long，所以用0x3f3f3f3f太小了 7 using namespace std; 8 9 int n,m; 10 11 struct edge{ 12 int to, cost; 13 edge(){} 14 edge(int to, int cost) : to(to), cost(cost){} 15 }; 16 17 vector <edge> ch[MAX]; 18 typedef pair<long long, int> P;//length,point 19 long long d[MAX];//2*10e5 * 10e5 会超出int 20 int road[MAX]; 21 int pre[MAX]; 22 23 24 void dijkstra(){ 25 priority_queue<P, vector<P>, greater<P> > que; 26 for(int i = 0; i <=n; i++) 27 d[i] = INF; 28 d[1] = 0; 29 que.push(P(0,1)); 30 31 while(que.size()) 32 { 33 if(que.empty()) break; 34 P p = que.top(); que.pop(); 35 int k = p.second; 36 if(d[k] < p.first) 37 continue; 38 for(int i = 0; i < ch[k].size(); i++) 39 { 40 edge e = ch[k][i]; 41 if(d[e.to] > d[k] + e.cost) 42 { 43 d[e.to] = d[k] + e.cost; 44 que.push(P(d[e.to],e.to)); 45 pre[e.to] = k; 46 } 47 } 48 } 49 } 50 51 52 int main() 53 { 54 int a,b,c; 55 cin >> n >> m; 56 for(int i = 0; i < m; i++) 57 { 58 cin >> a >> b >> c; 59 ch[a].push_back(edge(b,c)); 60 ch[b].push_back(edge(a,c)); 61 } 62 dijkstra(); 63 if(d[n] == INF) 64 { 65 cout<<"-1 "; 66 return 0; 67 } 68 int now = n, cou = 0; 69 while(now != 1) 70 { 71 road[cou++] = now; 72 now = pre[now]; 73 } 74 road[cou++] = 1; 75 for(int i = cou-1; i >= 0; i--) 76 cout<< road[i]<<" "; 77 return 0; 78 }
Sorting
1 #include <iostream> 2 #include <algorithm> 3 using namespace std; 4 5 int main() 6 { 7 int n,ch[100010]; 8 cin >> n; 9 for(int i = 0; i < n; i++) 10 cin >>ch[i]; 11 sort(ch,ch+n); 12 int ans = 0; 13 for(int i = 0; i < n; i++) 14 { 15 if(ch[i] > ans) 16 ans++; 17 } 18 cout << ans+1 <<endl; 19 return 0; 20 } 21 /*在非递减序列中，求不能连续到达的最小的数 22 如果数组元素的值比下标大，那么数组可以减小 23 24 25 */