标签:
input | output |
---|---|
5 1234567 666666 3141593 666666 4343434 5 1 5 3141593 1 5 578202 2 4 666666 4 4 7135610 1 1 1234567 |
10101
|
分析:map+二分;
代码:
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorithm> #include <climits> #include <cstring> #include <string> #include <set> #include <map> #include <queue> #include <stack> #include <vector> #include <list> #define rep(i,m,n) for(i=m;i<=n;i++) #define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++) #define mod 1000000007 #define inf 0x3f3f3f3f #define vi vector<int> #define pb push_back #define mp make_pair #define fi first #define se second #define ll long long #define pi acos(-1.0) const int maxn=1e5+10; const int dis[4][2]={{0,1},{-1,0},{0,-1},{1,0}}; using namespace std; ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);} ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;} int n,m,k,t; string ans; map<int,set<int> >a; set<int>::iterator it; int main() { int i,j; scanf("%d",&n); rep(i,1,n)scanf("%d",&j),a[j].insert(i); int q; scanf("%d",&q); while(q--) { int l,r,p; scanf("%d%d%d",&l,&r,&p); if((it=a[p].lower_bound(l))!=a[p].end()&&*it<=r) ans+=‘1‘; else ans+=‘0‘; } cout<<ans<<endl; //system("pause"); return 0; }
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原文地址:http://www.cnblogs.com/dyzll/p/5785154.html