hdu-5742 It's All in the Mind (math)

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It's all on the mind

Time limit:2000/1000 MS (java/others)

Memory limit:65536/65536 K (java/others)

problem DescriptionProfessor Zhang has a number sequencea1,a2,.. . ,an . However, the sequence isn't complete and some elements is missing. Fortunately, Professor Zhang remembers some properties of the sequence:

1. For everyi∈{1,2,.. . ,N} ,0≤ai≤ .
2. The sequence is non-increasing, i.e.a1≥a2≥. . ≥an .
3. The sum of all elements in the sequence are not zero.

Professor Zhang wants to know the maximum value ofA1+ a2 ∑n i =1ai Among all the possible sequences.

InputThere is multiple test cases. The first line of input contains an integerT, indicating the number of test cases. For each test case:

The first contains and the integersNandm (2≤n≤,0≤m≤n) -The length of the sequence and the number of known elements.

In the nextmLines, each contains, integersxi andyi (1≤XI≤N,0≤YI≤100,XI<xi+1,yi≥yi+1) , indicating that axi=yi.

OutputFor each test case, the output of the answer as an irreducible fraction "P/Q", where P, q is integers, q>0.

Sample Input 22 03 13 1

Sample output 1/1200/201  Test Instructions:  gives a part of a monotone sequence that allows you to find the maximum value of this equation;  the minimum value of:  (A1+A2)/(∑ai); it is ∑ai/(a1+ A2) is the maximum value of 1+ (a3+a4+...an)/(A1+A2), then the numerator is as small as possible, the denominator of the largest, equivalent to the reduction of inequalities; &NBSP;AC Code:

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include < cmath> #include <bits/stdc++.h> #include <stack>using namespace std; #define for (i,j,n) for (int i=j;i< =n;i++) #define MST (SS,B) memset (ss,b,sizeof (ss)); typedef long LONG ll;template<class t> void Read (T&num) {C Har CH;    BOOL F=false; For (Ch=getchar (); ch< ' 0 ' | | Ch> ' 9 ';    f= ch== '-', Ch=getchar ()); for (num=0;    ch>= ' 0 ' &&ch<= ' 9 '; num=num*10+ch-' 0 ', Ch=getchar ()); F && (num=-num);}    int stk[70], tp;template<class t> inline void print (T p) {if (!p) {puts ("0"); return;}    while (p) stk[++ TP] = p%10, p/=10;    while (TP) Putchar (stk[tp--] + ' 0 '); Putchar (' \ n ');} Const LL Mod=1e9+7;const double Pi=acos ( -1.0); const int INF=1E9;CONST int N=2e5+10;const int Maxn=500+10;const double eps= 1e-6;int A[MAXN];     int gcd (int x,int y) {if (y==0) return x; return gcd (y,x%y);        }int Main () {int t;        Read (t); While(t--)            {int n,m;            Read (n); read (m);            MST (A,-1);            int x, y;                for (i,1,m) {read (x); Read (y);            A[x]=y;            } int sum=0;            if (a[n]==-1) a[n]=0;            Sum+=a[n];                for (int i=n-1;i>2;i--) {if (a[i]==-1) a[i]=a[i+1];            Sum+=a[i];            } if (a[1]==-1) a[1]=100;            if (a[2]==-1) a[2]=a[1];            int p,q;            P=A[1]+A[2];            Q=sum+p;            if (q==0) int g=gcd (P,Q);        cout<<p/g<< "/" <<q/g<<endl; } return 0;}

　　

hdu-5742 It's All in the Mind (math)