题目传送门:https://codeforces.com/problemset/problem/600/B
题目大意: 给定长度为\(n\)的序列\(A\)和长度为\(m\)的序列\(B\),对于\(B_i\)找到\(A_j\leqslant B_i\)的个数
排序,二分
乱搞就完事
/*program from Wolfycz*/ #include #include #include #include #include #include #include #define Fi first #define Se second #define ll_inf 1e18 #define MK make_pair #define sqr(x) ((x)*(x)) #define pii pair #define int_inf 0x7f7f7f7f using namespace std; typedef long long ll; typedef unsigned int ui; typedef unsigned long long ull; inline char gc(){ static char buf[1000000],*p1=buf,*p2=buf; return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++; } templateinline T frd(T x){ int f=1; char ch=gc(); for (;ch<'0'||ch>'9';ch=gc()) if (ch=='-') f=-1; for (;ch>='0'&&ch<='9';ch=gc()) x=(x<<1)+(x<<3)+ch-'0'; return x*f; } templateinline T read(T x){ int f=1; char ch=getchar(); for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1; for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-'0'; return x*f; } inline void print(int x){ if (x<0) putchar('-'),x=-x; if (x>9) print(x/10); putchar(x%10+'0'); } const int N=2e5; int A[N+10]; int main(){ // freopen(".in","r",stdin); // freopen(".out","w",stdout); int n=read(0),m=read(0); for (int i=1;i<=n;i++) A[i]=read(0); sort(A+1,A+1+n); for (int i=1;i<=m;i++){ int pos=upper_bound(A+1,A+1+n,read(0))-A; printf("%d%c",pos-1,i==m?'\n':' '); } return 0; }