题意:略
先给a排序。然后做差记在b数组,b排序,记录b数组前缀和。询问时利用了等长(记询问区间宽度w)的性质,重叠部分不相交。那么做差<w的两个位置相交,>w的不相交。然后发现不交的部分和前缀和有关。
#include <iostream> #include <cstdio> #include <cmath> #include <algorithm> #include <vector> #include <iomanip> #include <cstring> #include <map> #include <queue> #include <set> #include <cassert> #include <stack> #include <bitset> #define mkp make_pair #define err cout<<"err"<<endl using namespace std; const double EPS=1e-8; typedef long long lon; typedef unsigned long long ull; typedef pair<int,int> pii; const lon SZ=100010,SSZ=SZ*SZ,APB=4,one=1; const lon INF=0x3f3f3f3f,mod=1000000007; lon n,a[SZ],b[SZ],sum[SZ]; void init() { cin>>n; for(lon i=1;i<=n;++i)cin>>a[i]; sort(a+1,a+1+n); for(lon i=1;i<=n-1;++i) { b[i]=a[i+1]-a[i]; } sort(b+1,b+1+n-1); for(lon i=1;i<=n-1;++i) { sum[i]=sum[i-1]+b[i]; } lon qnum; cin>>qnum; for(lon tim=1;tim<=qnum;++tim) { lon ll,rr; cin>>ll>>rr; lon w=rr-ll+1; lon res=0; lon pos=lower_bound(b+1,b+1+n-1,w)-1-b; res=w+sum[pos]+(n-1-pos)*w; cout<<res<<endl; } } void work() { } void release() { } int main() { std::ios::sync_with_stdio(0); //freopen("d:\\1.txt","r",stdin); lon casenum; //cin>>casenum; //cout<<casenum<<endl; //for(lon tim=1;tim<=casenum;++tim) //for(lon tim=1;cin>>n;++tim) { //cout<<"Case #"<<tim<<": "; init(); work(); release(); } return 0; }
codeforces 1119 D. Frets On Fire
原文:https://www.cnblogs.com/gaudar/p/11551498.html