打div3翻车了
A.第一个操作是除二，第二个操作视为两下操作之后除三，第三个操作视为三下操作之后除五，直接计算贡献
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x); #define Pri(x) printf("%d ", x) #define Prl(x) printf("%lld ",x); #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();} while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;} const double eps = 1e-9; const int maxn = 110; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; LL N; int main(){ int T; Sca(T); while(T--){ Scl(N); int ans = 0; while(!(N % 2)){ N /= 2; ans++; } while(!(N % 3)){ N /= 3; ans += 2; } while(!(N % 5)){ N /= 5; ans += 3; } if(N != 1) ans = -1; Pri(ans); } return 0; }
B.毫无疑问先把所有数模3，整除的直接计入贡献，余1的找余2的凑一对计入贡献，最后剩下的三个凑一对计入贡献
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x); #define Pri(x) printf("%d ", x) #define Prl(x) printf("%lld ",x); #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();} while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;} const double eps = 1e-9; const int maxn = 1010; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; int a[maxn]; int b[maxn]; int main(){ int T; Sca(T); while(T--){ Sca(N); b[0] = b[1] = b[2] = 0; for(int i = 1; i <= N ; i ++){ Sca(a[i]); a[i] %= 3; b[a[i]]++; } if(b[1] > b[2]) swap(b[1],b[2]); Pri(b[0] + b[1] + (b[2] - b[1]) / 3); } return 0; }
C.比较套路的一个寻找子序列数量，开6个计数器即可。最后答案就是字符串长度len - 匹配成功的子序列数量 * 6
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x); #define Pri(x) printf("%d ", x) #define Prl(x) printf("%lld ",x); #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();} while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;} const double eps = 1e-9; const int maxn = 5e5 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; int a[maxn]; const int b[10] = {0,4,8,15,16,23,42}; int dp[maxn]; int main(){ Sca(N); for(int i = 1; i <= N ; i ++) Sca(a[i]); dp[0] = INF; for(int i = 1; i <= N ; i ++){ for(int j = 1; j <= 6; j ++){ if(b[j] == a[i] && dp[j - 1]){ dp[j - 1]--; dp[j]++; } } } Pri(N - dp[6] * 6); return 0; }
D.将所有数从大到小排序然后扫描，遇到的当前最大的数如果是素数说明是一个小素数变过来的，否则说明他要变成他最大的因子。
直接递推即可。
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x); #define Pri(x) printf("%d ", x) #define Prl(x) printf("%lld ",x); #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();} while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;} const double eps = 1e-9; const int maxn = 3e6 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; int a[maxn]; int prime[maxn]; int pos[maxn]; bool isprime[maxn]; void init(){ for(int i = 2 ; i <= 2750131; i ++) isprime[i] = 1; int cnt = 0; for(int i = 2 ; i <= 2750131; i ++){ if(!isprime[i]) continue; prime[++cnt] = i; pos[i] = cnt; for(int j = i + i; j <= 2750131; j += i) isprime[j] = 0; } } int delnum[maxn]; int find(int x){ for(int i = 2; i <= x; i ++){ if(!(x % i)) return max(i,x / i); } return 1; } int main(){ Sca(N); init(); for(int i = 1; i <= N * 2 ; i ++) Sca(a[i]); sort(a + 1,a + 1 + N * 2); for(int i = N * 2; i >= 1; i --){ if(delnum[a[i]]){ delnum[a[i]]--; continue; } if(isprime[a[i]]){ delnum[pos[a[i]]]++; printf("%d ",pos[a[i]]); }else{ int n = find(a[i]); delnum[n]++; printf("%d ",a[i ]); } } return 0; }
E.一看是个最小点覆盖，仔细看看发现没有要求最小，只是找一个点覆盖。
对原图进行一手黑白染色，然后白点集合和黑点集合必定有一个符合答案，取集合元素数量较小的那个集合输出。
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x); #define Pri(x) printf("%d ", x) #define Prl(x) printf("%lld ",x); #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();} while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;} const double eps = 1e-9; const int maxn = 2e5 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; struct Edge{ int to,next; }edge[maxn * 2]; int head[maxn],tot; int color[maxn]; vector<int>w[2]; void init(){ for(int i = 0 ; i <= N ; i ++){ color[i] = head[i] = -1; } w[0].clear(); w[1].clear(); tot = 0; } void add(int u,int v){ edge[tot].to = v; edge[tot].next = head[u]; head[u] = tot++; } void dfs(int t){ for(int i = head[t]; ~i ; i = edge[i].next){ int v = edge[i].to; if(color[v] != -1) continue; color[v] = 1 ^ color[t]; dfs(v); } } int main(){ int T; Sca(T); while(T--){ Sca2(N,M); init(); for(int i = 1; i <= M ; i ++){ int u,v; Sca2(u,v); add(u,v); add(v,u); } for(int i = 1; i <= N ; i ++){ if(color[i] == -1){ color[i] = 0; dfs(i); } } for(int i = 1; i <= N ; i ++) w[color[i]].push_back(i); if(w[0].size() > w[1].size()) swap(w[0],w[1]); Pri(w[0].size()); for(int i = 0 ; i < w[0].size(); i ++){ printf("%d ",w[0][i]); } puts(""); } return 0; }