A:sb题,多特判几下就好了
//#pragma comment(linker, "/stack:200000000") //#pragma GCC optimize("Ofast,no-stack-protector") //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") //#pragma GCC optimize("unroll-loops") #include<bits/stdc++.h> #define fi first #define se second #define mp make_pair #define pb push_back #define pi acos(-1.0) #define ll long long #define mod 1000000007 #define C 0.5772156649 #define ls l,m,rt<<1 #define rs m+1,r,rt<<1|1 #define pil pair<int,ll> #define pii pair<int,int> #define ull unsigned long long #define base 1000000000000000000 #define fio ios::sync_with_stdio(false);cin.tie(0) using namespace std; const double g=10.0,eps=1e-9; const int N=100000+10,maxn=3000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f; int main() { int x,y; scanf("%d%d",&x,&y); if(y==1) { if(x!=0)puts("No"); else puts("Yes"); return 0; } if(x>=y-1&&(x-y+1)%2==0&&y!=0)puts("Yes"); else puts("No"); return 0; } /******************** ********************/
B:n^2模拟
//#pragma comment(linker, "/stack:200000000") //#pragma GCC optimize("Ofast,no-stack-protector") //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") //#pragma GCC optimize("unroll-loops") #include<bits/stdc++.h> #define fi first #define se second #define mp make_pair #define pb push_back #define pi acos(-1.0) #define ll long long #define mod 1000000007 #define C 0.5772156649 #define ls l,m,rt<<1 #define rs m+1,r,rt<<1|1 #define pil pair<int,ll> #define pii pair<int,int> #define ull unsigned long long #define base 1000000000000000000 #define fio ios::sync_with_stdio(false);cin.tie(0) using namespace std; const double g=10.0,eps=1e-9; const int N=100000+10,maxn=3000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f; int main() { int n,ans=0; scanf("%d",&n); for(int i=1;i<=n;i++) { for(int j=i;j<=n;j++) { int te=i^j; if(j<=te&&te<=n&&i+j>te&&i+te>j&&j+te>i)ans++; } } printf("%d ",ans); return 0; } /******************** ********************/
C:找从1到k中,有没有n%i相同的情况
直接暴力,因为打表出来的结果k不超过15
//#pragma comment(linker, "/stack:200000000") //#pragma GCC optimize("Ofast,no-stack-protector") //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") //#pragma GCC optimize("unroll-loops") #include<bits/stdc++.h> #define fi first #define se second #define mp make_pair #define pb push_back #define pi acos(-1.0) #define ll long long #define mod 1000000007 #define C 0.5772156649 #define ls l,m,rt<<1 #define rs m+1,r,rt<<1|1 #define pil pair<int,ll> #define pii pair<int,int> #define ull unsigned long long #define base 1000000000000000000 #define fio ios::sync_with_stdio(false);cin.tie(0) using namespace std; const double g=10.0,eps=1e-9; const int N=100000+10,maxn=3000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f; map<ll,ll>vis; int main() { ll n,k; scanf("%lld%lld",&n,&k); // if(k>=n)puts("No"); // else { bool ok=1; for(ll i=1;i<=k;i++) { if(vis[n%i]) { ok=0; break; } else vis[n%i]=1; } if(ok)puts("Yes"); else puts("No"); } return 0; } /******************** ********************/
D:给n个字符串,要求重拍后组成的字符串,价值最大,价值是满足i<j,s[i]=='s'&&s[j]=='h'的对数
做法:直接sort,判断条件是前后组合一下,看那种情况的价值最大,求价值直接后缀和h,扫一遍遇到s就加上后缀,复杂度O(nlogn)
//#pragma comment(linker, "/stack:200000000") //#pragma GCC optimize("Ofast,no-stack-protector") //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") //#pragma GCC optimize("unroll-loops") #include<bits/stdc++.h> #define fi first #define se second #define mp make_pair #define pb push_back #define pi acos(-1.0) #define ll long long #define mod 1000000007 #define C 0.5772156649 #define ls l,m,rt<<1 #define rs m+1,r,rt<<1|1 #define pil pair<int,ll> #define pii pair<int,int> #define ull unsigned long long #define base 1000000000000000000 #define fio ios::sync_with_stdio(false);cin.tie(0) using namespace std; const double g=10.0,eps=1e-9; const int N=100000+10,maxn=3000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f; map<ll,ll>vis; int main() { ll n,k; scanf("%lld%lld",&n,&k); // if(k>=n)puts("No"); // else { bool ok=1; for(ll i=1;i<=k;i++) { if(vis[n%i]) { ok=0; break; } else vis[n%i]=1; } if(ok)puts("Yes"); else puts("No"); } return 0; } /******************** ********************/
E:n颗数,每个树ci个鸟,从1开始依次移动,每次可以召唤1<=i<=ci只鸟,每召唤一只鸟需要costi法力值,每召唤一只鸟法力值上限加上b,每移到一颗树法力加x,要求找到最大能召唤的鸟个数
做法:dp[i][j]维护的是当访问到第i颗树时,召唤了j只鸟的当前最大法力值,初始状态dp[0][0]=w,转移方程dp[i][j]=max(dp[i][j],min(dp[i-1][j-k]+x,w+(j-k)*b)-cost[i]*k)
最后dp[n]里最大能访问到的地方就是答案
//#pragma comment(linker, "/stack:200000000") //#pragma GCC optimize("Ofast,no-stack-protector") //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") //#pragma GCC optimize("unroll-loops") #include<bits/stdc++.h> #define fi first #define se second #define mp make_pair #define pb push_back #define pi acos(-1.0) #define ll long long #define mod 1000000007 #define C 0.5772156649 #define ls l,m,rt<<1 #define rs m+1,r,rt<<1|1 #define pil pair<int,ll> #define pii pair<int,int> #define ull unsigned long long #define base 1000000000000000000 #define fio ios::sync_with_stdio(false);cin.tie(0) using namespace std; const double g=10.0,eps=1e-9; const int N=1000+10,maxn=3000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f; ll dp[N][10000+10]; ll c[N],cost[N]; int main() { ll n,w,b,x,sum=0; scanf("%lld%lld%lld%lld",&n,&w,&b,&x); for(int i=1;i<=n;i++) { scanf("%lld",&c[i]); sum+=c[i]; } for(int i=1;i<=n;i++)scanf("%lld",&cost[i]); memset(dp,-1,sizeof dp); dp[0][0]=w; for(int i=1;i<=n;i++) { for(int j=0;j<=sum;j++) { for(int k=0;k<=min((ll)j,c[i]);k++) { if(dp[i-1][j-k]==-1)continue; // if(min(dp[i-1][j-k]+x,w+b*(j-k))-cost[i]*k<0)continue; dp[i][j]=max(dp[i][j],min(dp[i-1][j-k]+x,w+b*(j-k))-cost[i]*k); } } } int ans=0; while(ans+1<=sum&&dp[n][ans+1]!=-1)ans++; printf("%d ",ans); return 0; } /******************** ********************/
F:给你n,k要求构造一个集合,满足集合里所有数不超过n,没有重复,满足倍数关系的对数有k对
做法:先求出1到n的满足倍数关系的对数e[i],如果e[n]<k,则1到n每个都有也不能满足条件输出no,否则yes,贪心的在1到n删数来满足条件,从1到n每次删除之后在对应的倍数因子数d[i]减一(至于这样贪心为什么能满足条件,我也不会证明,直觉上可以),复杂度O(nlogn)
//#pragma comment(linker, "/stack:200000000") //#pragma GCC optimize("Ofast,no-stack-protector") //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") //#pragma GCC optimize("unroll-loops") #include<bits/stdc++.h> #define fi first #define se second #define mp make_pair #define pb push_back #define pi acos(-1.0) #define ll long long #define mod 1000000007 #define C 0.5772156649 #define ls l,m,rt<<1 #define rs m+1,r,rt<<1|1 #define pil pair<int,ll> #define pii pair<int,int> #define ull unsigned long long #define base 1000000000000000000 #define fio ios::sync_with_stdio(false);cin.tie(0) using namespace std; const double g=10.0,eps=1e-9; const int N=300000+10,maxn=3000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f; int d[N],e[N]; int main() { int n,k; scanf("%d%d",&n,&k); for(int i=1;i<=n;i++) for(int j=2*i;j<=n;j+=i) d[j]++; for(int i=1;i<=n;i++)e[i]=e[i-1]+d[i]; if(e[n]<k)puts("No"); else { vector<int>ans; int te=e[n]-k; for(int i=1;i<=n;i++) { if(te>=d[i]+n/i-1&&d[i]==1) { te-=d[i]+n/i-1; for(int j=2*i;j<=n;j+=i)d[j]--; } else ans.pb(i); } sort(ans.begin(),ans.end()); puts("Yes"); printf("%d ",ans.size()); for(int i=0;i<ans.size();i++)printf("%d ",ans[i]); puts(""); } return 0; } /******************** ********************/