Summer warm up B. Download speed

Recently, Nono finally ended the annual 12-month hibernation, the first thing to wake up--watch the movie!! Nono found that there have been a lot of good and powerful films in the last year, such as this, this, and this. So Nono directly threw all these movies into the download list. Known Nono downloaded n movies, of which part I movie size si, the movie maximum download speed will not exceed VI, at any time the download speed of all movies will not exceed Vt. Now Nono want to know how long it will take at least to get all of these movies under these conditions. Input multiple sets of inputs. The first behavior of each set of inputs is two integers n,vt (0<n≤106, 0<vt≤109). The next n rows of two integers per line represent Si, Vi (0<si,vi≤1000). The input ends with EOF. Output

For each set of data, output an approximate fraction of the minimum time required to download a full movie.

Sample Input

3 67 45 23 41 106 2

Sample Output

5/23/1
The problem begins with complexity, and all tasks start at the same time, and the maximum time to compare a single task.

The first version of the test all the data are no problem, do not know where the wrong is always WA

#include <stdio.h>structdongser{intA; intb;};structDongser D;intda,db;voidSimintIintj) {    intk=i-1; if(I==J) {i=j=1;} if(i%j==0) {i=i/j;j=1;d a=i;db=j;return;} if(j%i==0) {j=j/i;i=1;d a=i;db=j;return;}  while(1)    {        if(k==1) {da=i;db=j;return;} if(i%k==0&&j%k==0) {i=i/k;j=j/K;} Else{k--;} }}intMain () {intI,J,K,N,VT; Doublem,t,p;  while(SCANF ("%d%d", &AMP;N,&AMP;VT)! =EOF) {         for(i=1; i<=n;i++) {scanf ("%d%d",&d.a,&d.b); J+=d.a; T=d.a/d.b; if(t>p) {p=t;k=i;da=d.a;db=d.b;} } m=j/VT; if(m>=p) {Da=j;db=vt;sim (da,db);p rintf ("%d/%d\n", da,db);} if(m<p) {SIM (da,db);p rintf ("%d/%d\n", da,db);} I=j=k=n=vt=m=t=p=da=db=d.a=d.b=0; }    return 0;}

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AC Version:

#include <iostream>#include<cstdio>using namespacestd;intN, VT;intGCD (intAintb) {  while(b! =0){ intT a%= b; t = A; A = b; b =t;}returnA;}intMain () {Ios::sync_with_stdio (false);  while(Cin >> N >>VT) { intsum =0;intrs =0, RV =1;  for(inti =0; I < n; ++i) {ints, v; CIN>> s >> v; sum + = s;if((Double) Rs/rv < (Double) s/v) {rs = s; rv = V;}}if((Double) Rs/rv < (Double) (SUM/VT) {rs = sum; RV = VT;}intK = gcd (rs, RV); cout << rs/k <<"/"<< rv/k << Endl; }return 0; }

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Summer warm up B. Download speed