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41 Group the People Given the Group Size They Belong To

There are n people whose IDs go from 0 to n - 1 and each person belongs exactly ...

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解决Error:All flavors must now belong to a named flavor dimension.

提示:Error:All flavors must now belong to a named flavor dimension.Learn more at https://d.android.com/ To make this work, the plugin now requires that all flavors belong to a named flavor dimension —even Otherwise, you will get the following build error: Error:All flavors must now belong to a named flavor

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    Error:All flavors must now belong to a named flavor dimension. Learn more at https:d.android.com

    23810

    14:Challenge 7(map大法好)

    left[belong[a]]=a; 17 for (int a=1;a<=n;a++) 18 sum[belong[a]]+=z[a]; 19 20 int query(int l,int ans+=z[a]+col[belong[a]]; 29 for (int a=belong[l]+1;a<belong[r];a++) 30 ans+=sum[ a]; 31 for (int a=left[belong[r]];a<=r;a++) 32 ans+=z[a]+col[belong[a]]; 33 } 34 return ans; 35 } 36 37 void modify(int l,int r,int v) { 38 if (belong[l]==belong[r]) { 39 sum[belong[a]]+=v; 48 } 49 for (int a=belong[l]+1;a<belong[r];a++) { 50

    47240

    knn之构造kd树和最近邻求取c++实现

    ) {     node *nearest=belong;      double mindis=distance(key,belong);      //cout<<mindis<<" mindis * belong=find_first_belong(key,root);     //cout<<belong->x.first<<" "<<belong->x.second<<endl;     ,node*root) {     node *nearest=belong;      double mindis=distance(key,belong);      //cout<<belong      {      //cout<<belong->x.first<<" "<<belong->x.second<<endl;      node*fat=belong->father;       * belong=find_first_belong(key,root);     //cout<<belong->x.first<<" "<<belong->x.second<<endl;

    20800

    LOJ#6279. 数列分块入门 3

    [l]); if(belong[l]! =belong[r]) { for(int i=L[r];i<=r;i++) a[i]+=val; Sort(belong[r]); } (ans,a[i]+tag[belong[l]]); if(belong[l]! +tag[belong[r]]); for(int i=belong[l]+1;i<=belong[r]-1;i++) { int x=val-tag[i]; [i]=(i-1)/block+1,L[i]=(belong[i]-1)*block+1,R[i]=belong[i]*block; for(int i=1;i<=N;i++) v[belong

    41190

    LOJ#6280. 数列分块入门 4

    [i]]+=val; if(belong[l]! =belong[r]) for(int i=L[r];i<=r;i++) a[i]+=val,sum[belong[i]]+=val; for(int i =belong[l]+1;i<=belong[r]-1;i++) tag[i]+=val; } int Query(int l,int r,int mod) { int ans= =belong[r]) for(int i=L[r];i<=r;i++) ans+=a[i]+tag[belong[i]]; for(int i=belong [i]=(i-1)/block+1,L[i]=(belong[i]-1)*block+1,R[i]=belong[i]*block; for(int i=1;i<=N;i++) a[i]=read

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    P3119 草鉴定Grass Cownoisseur 缩点 topo或最长路

    输出 ans - sum[belong[1]] ,因为这个重复了。 [1]] = sum[belong[1]]; //初始化 防止走不通 f[i] 表示到第 i 块联通分量的最大数目 for(int i=1;i<=n;i++){ if(! =belong[v]){ pos.add(belong[i],belong[v]); neg.add(belong[v],belong[i]); } } } pos.to_Sort v,j=head[i];j;j=edge[j].nxt){ v = edge[j].v; if(pos.f[belong[v]]>0&&neg.f[belong[i]]>0)//两点都必须到达 ans = max(ans,pos.f[belong[v]]+neg.f[belong[i]]); } } printf("%d\n",ans-sum[belong[1]]); return

    15610

    LOJ#6281. 数列分块入门 5

    [i]]-=a[i]; a[i]=sqrt(a[i]); sum[belong[i]]+=a[i]; } if(belong[l]! =belong[r]) for(int i=L[r];i<=r;i++) sum[belong[i]]-=a[i],a[i]=sqrt(a[i]),sum[belong[i]]+=a[i]; for(int i=belong[l]+1;i<=belong[r]-1;i++) { if(flag[i]) {continue;} flag[i]=1; for(int j=L[i* =belong[r]) for(int i=L[r];i<=r;i++) ans+=a[i]; for(int i=belong[l]+1;i<=belong[r]-1;i++) ans [i]=(i-1)/block+1,L[i]=(belong[i]-1)*block+1,R[i]=belong[i]*block; for(int i=1;i<=N;i++) sum[belong[

    455110

    LOJ#6283. 数列分块入门 7

    =belong[r]) { reset(belong[r]); for(int i=L[r];i<=r;i++) a[i]+=val,a [i]%=mod; } for(int i=belong[l]+1;i<=belong[r]-1;i++) add[i]+=val,add[i]%=mod; } void =mod; if(belong[l]! =belong[r]) { reset(belong[r]); for(int i=L[r];i<=r;i++) a[i]*=val,a[ i]%=mod; } for(int i=belong[l]+1;i<=belong[r]-1;i++) mul[i]*=val,add[i]*=val

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    LOJ#6278. 数列分块入门 2

    [l]); if(belong[l]! =belong[r]) { for(int i=L[r];i<=r;i++) a[i]+=val; Sort(belong[r]); } for(int i=belong[l]+1;i<=belong[r]-1;i++) tag[i]+=val; } int Query(int l,int r,int val) { int ans=0; for(int i=l;i<=min(r,R[l]);i++) if(a[i]+tag[belong[l]]<val) ans++; if(belong [i]=(i-1)/block+1,L[i]=(belong[i]-1)*block+1,R[i]=belong[i]*block; for(int i=1;i<=N;i++) v[belong

    42050

    洛谷P3245 大数(莫队)

    [l] == belong[rhs.l] ? r < rhs.r : belong[l] < belong[rhs.l]; } } q[MAXN]; void Add(int x) { now += cnt[x]; cnt[ [l] == belong[rhs.l] ? [l] == belong[rhs.l] ? r < rhs.r : belong[l] < belong[rhs.l]; } }q[MAXN]; void Add(int x, int opt, int pos) { cnt[x]

    22710

    LOJ#6277. 数列分块入门 1

    ;c=nc();} while(c>='0'&&c<='9'){x=x*10+c-'0';c=nc();} return x*f; } int N,a[MAXN],tag[MAXN],belong =belong[r]) for(int i=L[r];i<=r;i++) a[i]+=val; for(int i=belong[l]+1;i<=belong[r]-1;i++) tag[i]+ [i]=(i-1)/block+1; for(int i=1;i<=N;i++) L[i]=(belong[i]-1)*block+1; for(int i=1;i<=N;i++) R[ i]=belong[i]*block; for(int i=1;i<=N;i++) { int opt=read(),l=read(),r=read(),c=read() ; if(opt==0) IntervalAdd(l,r,c); else printf("%d\n",a[r]+tag[belong[r]]); } return

    38460

    LOJ#6284. 数列分块入门 8

    (c=='-')f=-1;c=nc();} while(c>='0'&&c<='9'){x=x*10+c-'0';c=nc();} return x*f; } int a[MAXN],belong ++) a[i]=tag[x]; tag[x]=-1; } int Query(int l,int r,int val) { int ans=0; reset(belong =belong[r]) { reset(belong[r]); for(int i=L[r];i<=r;i++) { if (a[i]==val) ans++; a[i]=val; } } for(int i=belong[l]+1;i<=belong[r]-1;i++ [i]=(i-1)/block+1,L[i]=(belong[i]-1)*block+1,R[i]=belong[i]*block; for(int i=1;i<=N;i++) {

    387100

    codechef QCHEF(不删除莫队)

    '0', c = getchar(); return x * f; } int N, K, M, a[MAXN], l[MAXN], r[MAXN], tl[MAXN], tr[MAXN], belong [l] == belong[rhs.l] ? r < rhs.r : belong[l] < belong[rhs.l]; } }; vector<query> q[MAXN]; int solve(int l, int r) { [i] = (i - 1) / block + 1, chmax(mx, belong[i]); for(int i = 1; i <= M; i++) { int l = read (), r = read(); if(belong[l] == belong[r]) ans[i] = solve(l, r); else q[belong[l]].push_back

    22930

    BZOJ4939: 掉进兔子洞(莫队 bitset)

    '0', c = getchar(); return x * f; } int N, M; int a[MAXN], date[MAXN], L[MAXN][5], R[MAXN][5], belong struct Query { int l, r, opt, id; bool operator < (const Query &rhs) const { return belong [l] == belong[rhs.l] ? belong[r] < belong[rhs.r] : belong[l] < belong[rhs.l]; } }Q[MAXN * 3 + 1]; bitset<MAXN> bit[B + 50 read(); M = read(); base = sqrt(N); for(int i = 1; i <= N; i++) a[i] = read(), date[i] = a[i], belong

    21910

    强连通专题

    <int>edge[MN]; vector<int>SCC[MN]; stack<int>s; int low[MN],dfn[MN]; int instack[MN],stap[MN]; int belong if(low[x]==dfn[x]) { int top; do { top=stap[p]; belong [i]; int y=belong[edge[i][j]]; if(x! [i]; int y=belong[edge[i][j]]; if(x! [i]; int y=belong[edge[i][j]]; if(x!

    39780

    强联通模板

    <int>edge[MN]; vector<int>SCC[MN]; stack<int>s; int low[MN],dfn[MN]; int instack[MN],stap[MN]; int belong if(low[x]==dfn[x]) { int top; do { top=stap[p]; belong dfn)); memset(low,0,sizeof(low)); memset(instack,0,sizeof(instack)); memset(belong ,0,sizeof(belong)); memset(ind,0,sizeof(ind)); for(i=1;i<=n;i++) { [i]; int y=belong[edge[i][j]]; if(x!

    29850

    洛谷P3356 火星探险问题(费用流)

    [i][j],belong[i][j]+point,-1,1); if(opt! =1) AddEdge(belong[i][j],belong[i][j]+point,0,INF); if(i<N) AddEdge(belong[i][j]+point,belong [i+1][j],0,INF); if(j<M) AddEdge(belong[i][j]+point,belong[i][j+1],0,INF); } S=0;T=point*4; AddEdge(S,belong[1][1],0,K); AddEdge(belong[N][M]+point,T,0,K); MCMF(); for(int i=1;i<=N;i++) for(int j=1;j<=M;j++) for(int k=head[belong[i][j]];k!

    40540

    P3387 【模板】缩点 Tarjan+记忆化搜索

    ll long long using namespace std; const int N = 10005; const int M = 300005; int dfn[N],low[N],cnt,belong (book[v])low[u] = min(dfn[v],low[u]); //v在栈内 u能到达的边是v的序号和u能到达最小的序号中较小的 } if(dfn[u]==low[u]){ belong [u] = ++cnt; int v; do{ v = s.top(); book[v]=0; s.pop(); belong[v]=cnt; sum[cnt]+= dfn[i]){ tarjan(i); } } memset(head,0,sizeof(head)); for(int i=1;i<=m;i++){ if(belong[a[i =belong[b[i]]){ add(belong[a[i]],belong[b[i]]); } } int ans=0; for(int i=1;i<=cnt;i++){ if(

    27530

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