http://codeforces.com/contest/940
A水题
//#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 cd complex<double> #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-12; const int N=100000+10,maxn=1200000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f; int a[N]; int main() { int n,d; scanf("%d%d",&n,&d); for(int i=0;i<n;i++)scanf("%d",&a[i]); sort(a,a+n); int minn=100000; for(int i=0;i<n;i++) { bool ok=0; for(int j=i;j<n;j++) { if(a[j]-a[i]>d) { ok=1; // printf("%d %d ",i,j-1); minn=min(minn,n-(j-i)); break; } } if(!ok) { // printf("%d+++ ",i); minn=min(minn,i); } } printf("%d ",minn); return 0; } /******************** ********************/
B直接暴力，忘了考虑1 的情况t了一发
//#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 cd complex<double> #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-12; const int N=100000+10,maxn=1200000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f; int main() { ll n,k,a,b,ans=0; scanf("%lld%lld%lld%lld",&n,&k,&a,&b); if(k==1) { printf("%lld ",(n-1)*a); return 0; } while(n) { if(n<k) { ans+=a*(n-1); break; } else { if(n%k==0) { ans+=min((n-n/k)*a,b); n=n/k; } else { ans+=(n%k)*a; n-=n%k; } } } printf("%lld ",ans); return 0; } /******************** ********************/
C从后往前扫找第一个个能替换的换了，然后把后面的全部换成最小,O(n*26)
//#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 cd complex<double> #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-12; const int N=100000+10,maxn=1200000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f; char s[N]; int num[30]; int main() { int n,k; scanf("%d%d%s",&n,&k,s); for(int i=0;i<n;i++)num[s[i]-'a']=1; int f=0; for(int i=0;i<26;i++) { if(num[i]) { f=i; break; } } if(n<k) { printf("%s",s); for(int i=0;i<k-n;i++)printf("%c",(char)(f+'a')); puts(""); } else { for(int i=k-1;i>=0;i--) { int ok=-1; for(int j=0;j<26;j++) { if(num[j]&&s[i]-'a'<j) { // printf("%d %d ",i,j); ok=j; break; } } if(ok!=-1) { s[i]=(char)(ok+'a'); // printf("%d %c ",i,s[i]); for(int j=i+1;j<k;j++)s[j]=(char)(f+'a'); for(int j=0;j<k;j++)printf("%c",s[j]); break; } } } return 0; } /******************** ********************/
D枚举所有情况找l和r即可，见过最简单的d题= =
//#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 cd complex<double> #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-12; const int N=100000+10,maxn=1200000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f; char s[N]; int a[N],b[N]; int main() { int n; scanf("%d",&n); for(int i=1;i<=n;i++)scanf("%d",&a[i]); scanf("%s",s+1); for(int i=1;i<=n;i++) { if(s[i]=='1')b[i]=1; else b[i]=0; } int l=-1e9,r=1e9; for(int i=5;i<=n;i++) { if(!b[i]&&b[i-1]&&b[i-2]&&b[i-2]&&b[i-3]) { r=min(r,a[i]-1); r=min(r,a[i-1]-1); r=min(r,a[i-2]-1); r=min(r,a[i-3]-1); r=min(r,a[i-4]-1); } else if(b[i]&&!b[i-1]&&!b[i-2]&&!b[i-2]&&!b[i-3]) { l=max(l,a[i]+1); l=max(l,a[i-1]+1); l=max(l,a[i-2]+1); l=max(l,a[i-3]+1); l=max(l,a[i-4]+1); } } printf("%d %d ",l,r); return 0; } /******************** ********************/
E，题意：把一堆数分成若干块，每个块的价值是去掉前k/c（k是块大小）小的数的和，求所有块最小的和
很明显每个块分尽可能的c个，然后每次可以用dp来算dp【i】=min（dp【i-1】+a【i】，dp【i-c】+sum（i-c，i））求sum复杂度是o(n)，可以用线段树预处理然后优化到o（logn），总复杂度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 cd complex<double> #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-12; const int N=100000+10,maxn=1200000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f; ll mi[N<<2],sum[N<<2],a[N],dp[N]; void pushup(int rt) { mi[rt]=min(mi[rt<<1],mi[rt<<1|1]); sum[rt]=sum[rt<<1]+sum[rt<<1|1]; } void build(int l,int r,int rt) { if(l==r) { mi[rt]=sum[rt]=a[l]; return ; } int m=(l+r)>>1; build(ls);build(rs); pushup(rt); } ll querysum(int L,int R,int l,int r,int rt) { if(L<=l&&r<=R)return sum[rt]; int m=(l+r)>>1; ll ans=0; if(L<=m)ans+=querysum(L,R,ls); if(m<R)ans+=querysum(L,R,rs); return ans; } ll querymi(int L,int R,int l,int r,int rt) { if(L<=l&&r<=R)return mi[rt]; int m=(l+r)>>1; ll ans=1e9+10; if(L<=m)ans=min(ans,querymi(L,R,ls)); if(m<R)ans=min(ans,querymi(L,R,rs)); return ans; } int main() { int n,c; scanf("%d%d",&n,&c); for(int i=1;i<=n;i++)scanf("%lld",&a[i]); build(1,n,1); for(int i=1;i<=n;i++) { dp[i]=dp[i-1]+a[i]; if(i-c>=0) { ll all=querysum(i-c+1,i,1,n,1); dp[i]=min(dp[i],dp[i-c]+all-querymi(i-c+1,i,1,n,1)); } } printf("%lld ",dp[n]); return 0; } /******************** ********************/
F，题意：一些数，两个操作1，把某个位置的数改成另一个，2，查询mex（数的总和个数）
带修莫队裸题，先把数离散化，然后暴力跑莫队即可，维护一个数的个数，一个数的个数有多少个，注意块一定要开n^(2/3)个,sqrt(n)会t，复杂度O（n^(5/3)+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 cd complex<double> #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-12; const int N=300000+10,maxn=1200000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f; int a[N],belong[N],ans[N],now[N]; int l=1,r=0,num[N],t,co[N]; int Hash[N]; struct que{ int l,r,tim,id; }qu[N]; struct change{ int pos,suf,pre; }ch[N]; bool cmp(que a,que b) { if(belong[a.l]==belong[b.l]) { if(belong[a.r]==belong[b.r])return a.tim<b.tim; else return a.r<b.r; } else return a.l<b.l; } void gao(int x,int d) { if(d==1) { num[co[x]]--; co[x]++; num[co[x]]++; } else { num[co[x]]--; co[x]--; num[co[x]]++; } } void go(int x,int d) { if(l<=x&&x<=r)gao(d,1),gao(a[x],-1); a[x]=d; } int main() { int n,q,cnt=0; scanf("%d%d",&n,&q); int block=3000; for(int i=1;i<=n;i++) { scanf("%d",&a[i]); Hash[cnt++]=a[i]; now[i]=a[i]; belong[i]=(i-1)/block+1; } int cnt1=0,cnt2=0; for(int i=1;i<=q;i++) { int op,a,b; scanf("%d%d%d",&op,&a,&b); if(op==1)qu[++cnt1]={a,b,cnt2,cnt1}; else ch[++cnt2]={a,b,now[a]},now[a]=b,Hash[cnt++]=b; } sort(Hash,Hash+cnt); cnt=unique(Hash,Hash+cnt)-Hash; for(int i=1;i<=n;i++)a[i]=lower_bound(Hash,Hash+cnt,a[i])-Hash; for(int i=1;i<=cnt2;i++) { ch[i].pre=lower_bound(Hash,Hash+cnt,ch[i].pre)-Hash; ch[i].suf=lower_bound(Hash,Hash+cnt,ch[i].suf)-Hash; } // for(int i=1;i<=cnt1;i++) // printf("%d %d %d %d ",qu[i].l,qu[i].r,qu[i].id,qu[i].tim); sort(qu+1,qu+1+cnt1,cmp); for(int i=1;i<=cnt1;i++) { while(t<qu[i].tim)go(ch[t+1].pos,ch[t+1].suf),t++; while(t>qu[i].tim)go(ch[t].pos,ch[t].pre),t--; while(l<qu[i].l)gao(a[l],-1),l++; while(l>qu[i].l)gao(a[l-1],1),l--; while(r<qu[i].r)gao(a[r+1],1),r++; while(r>qu[i].r)gao(a[r],-1),r--; int res=1; while(num[res])res++; ans[qu[i].id]=res; } for(int i=1;i<=cnt1;i++)printf("%d ",ans[i]); return 0; } /******************** ********************/