实现方法
补一个STL的离散化。
先排序,再unique一下,然后lower_bound找出每个数再离散后序列里对应的位置。
代码
#include <cstdio>
#include <algorithm>
using namespace std;
namespace fast_IO{
const int IN_LEN = 10000000, OUT_LEN = 10000000;
char ibuf[IN_LEN], obuf[OUT_LEN], *ih = ibuf + IN_LEN, *oh = obuf, *lastin = ibuf + IN_LEN, *lastout = obuf + OUT_LEN - 1;
inline char getchar_(){return (ih == lastin) && (lastin = (ih = ibuf) + fread(ibuf, 1, IN_LEN, stdin), ih == lastin) ? EOF : *ih++;}
inline void putchar_(const char x){if(oh == lastout) fwrite(obuf, 1, oh - obuf, stdout), oh = obuf; *oh ++= x;}
inline void flush(){fwrite(obuf, 1, oh - obuf, stdout);}
int read(){
int x = 0; int zf = 1; char ch = ' ';
while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar_();
if (ch == '-') zf = -1, ch = getchar_();
while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar_(); return x * zf;
}
void write(int x){
if (x < 0) putchar_('-'), x = -x;
if (x > 9) write(x / 10);
putchar_(x % 10 + '0');
}
}
using namespace fast_IO;
const int MAXN = 100005;
int a[MAXN], b[MAXN];
int main(){
int n = read();
for(int i = 0; i < n; i++)
a[i] = b[i] = read();
sort(b, b + n);
int m = unique(b, b + n) - b;
for(int i = 0; i < n; i++)
a[i] = lower_bound(b, b + m, a[i]) - b;
return 0;
}