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题意:
长度n的序列, m个询问区间[L, R], 问区间内的所有子段的不同GCD值有多少种.
题解:
固定右端点
预处理出 每个点向左延伸 的 不同gcd值
这样的 值不会超过log a 个
然后问题就变成了 问你一段区间内不同 gcd 值有多少,值是很少的 (询问一个区间有多少颜色的题型)
树状数组维护就可以了
#include<iostream> #include<algorithm> #include<cstdio> #include<cstring> #include<cmath> #include<vector> using namespace std; #pragma comment(linker, "/STACK:102400000,102400000") #define ls i<<1 #define rs ls | 1 #define mid ((ll+rr)>>1) #define pii pair<int,int> #define MP make_pair typedef long long LL; const long long INF = 1e18; const double Pi = acos(-1.0); const int N = 1e6+10, M = 1e2+11, mod = 1e9+7, inf = 2e9; int n,q,a[N],ans[N]; int gcd(int a, int b) { return b == 0 ? a : gcd(b, a%b);} vector<pii> G[N]; struct QQ{ int l,r,id; bool operator < (const QQ &a) const { return a.r > r; } }Q[N]; int C[N],vis[N]; void update(int x,int c) { for(int i =x; i < N; i+=i&(-i)) C[i] += c; } int ask(int x) { int s =0 ; for(int i = x; i; i-= i & (-i))s += C[i]; return s; } int main() { while(scanf("%d%d",&n,&q)!=EOF) { for(int i = 1; i <= n; ++i) scanf("%d",&a[i]); for(int i = 0; i <= n; ++i) G[i].clear(); for(int i = 1; i <= n; ++i) { int x = a[i]; int y = i; for(int j = 0; j < G[i-1].size(); ++j) { int res = gcd(x,G[i-1][j].first); if(x != res) { G[i].push_back(MP(x,y)); x = res; y = G[i-1][j].second; } } G[i].push_back(MP(x,y)); } memset(C,0,sizeof(C)); memset(vis,0,sizeof(vis)); for(int i = 1; i <= q; ++i) {scanf("%d%d",&Q[i].l,&Q[i].r),Q[i].id = i;} sort(Q+1,Q+q+1); for(int R = 0, i = 1; i <= q; ++i) { while(R < Q[i].r) { R++; for(int j = 0; j < G[R].size(); ++j) { int res = G[R][j].first; int ids = G[R][j].second; if(vis[res]) update(vis[res],-1); vis[res] = ids; update(vis[res],1); } } ans[Q[i].id] = ask(R) - ask(Q[i].l-1); } for(int i = 1; i <= q; ++i) cout<<ans[i]<<endl; } return 0; }
HDU 5869 Different GCD Subarray Query 离线+树状数组
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原文地址:http://www.cnblogs.com/zxhl/p/5865948.html