AGC 016B. Colorful Hats (IDEA)

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For a discussion of an element's classification, it is possible to know the total number of colors of n elements.
For example, the a[1] category discussion:

If the color of 1 is the same as an element, the total number of colors is a[1]. A[i] is either equal to a[1] (i is the same color as an element, with a count of a), or equal to a[1]+1 (i is the only color, the number is B).
To satisfy: B is not equal to n-1 (must have an I and 1 colors the same); maximum number of colors = (a+1)/2+b is greater than or equal to a[1], and there is an upper bound limit! That is, the minimum number of colors =1+b to be less than or equal to a[1].

If the color of 1 is unique, the total number of colors is a[1]+1. A[i] is either equal to a[1] (the color of I is unique, count to a), or equal to a[1]+1 (the same color as an element, the number of counts is B).
To satisfy: B cannot be 1 (can be greater than 1 can also be 0); The maximum number of colors =1+A+B/2 is greater than or equal to a[1]+1, and there is an upper bound limit! That is, the minimum number of colors =1+a+ (b!=0) is less than or equal to a[1]+1.

Of course This classification discusses how many of the minimum values are actually written by others (only one check is performed). I'm not saying that I wrote a problem. In fact, it is similar.

2ms 896kb#include <cstdio> #include <cctype> #include <algorithm>//#define GC () GetChar () #define Maxin 300000#define GC () (ss==tt&& (tt= (ss=in) +fread (In,1,maxin,stdin), SS==TT)? eof:*ss++) typedef long LONG ll;const int n=1e6+5;int a[n];char in[maxin],*ss=in,*tt=in;inline int read () {int Now=0;reg    Ister Char C=GC (); for (;!    IsDigit (c); C=GC ());    for (; IsDigit (c); now=now*10+c-' 0 ', C=GC ()); return now;}    BOOL Check1 (const int n) {int a=0,b=0;        for (int i=2,a1=a[1]; i<=n; ++i) if (A[I]==A1) ++a;        else if (a[i]==a1-1) ++b;    else return 0; Return b<n-1&& (a+1>>1) +b>=a[1]&&b<a[1];//upper bound limit ah ah}bool Check2 (const int n) {int a=0,b=    0;        for (int i=2,a1=a[1]; i<=n; ++i) if (A[I]==A1) ++a;        else if (a[i]==a1+1) ++b;    else return 0; Return b!=1&&a+ (b>>1) >=a[1]&&a+ (b!=0) <=a[1];}    int main () {int n=read ();    for (int i=1; i<=n; ++i) A[i]=read (); Puts (Check1 (n)|| Check2 (n)? "    Yes ":" No "); return 0;}

AGC 016B. Colorful Hats (IDEA)