hdu-5747 Aaronson (water problem)

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Aaronson

Time limit:4000/2000 MS (java/others)

Memory limit:131072/131072 K (java/others)

Problem descriptionrecently, Peter saw the equationX0+2X1+4x2++ 2mx M=n . He wants to find a solution  (x 0, x1 , X2 , Xm) In such a manner that∑I=0mxi is minimum and every xi (0≤i≤m) is non-negative.

Inputthere is multiple test cases. The first line of input contains an integerT(1≤T≤5) , indicating the number of test cases. For each test case:

The first contains and the integersNand m (0≤n,m≤9) .

Outputfor each test case, output the minimum value of∑I=0mxi.

Sample Input101 23 25 210 210 310 413 520 411 1112 3

Sample Output1223223232 Test Instructions:

Divide n into such, coefficients and smallest, greedy; 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)    {char 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=1e5+10;const int Maxn=500+10;const double eps= 1e-8;    LL f[60];void Init () {f[0]=1;    for (i,1,50) {f[i]=f[i-1]*2;        }}int Main () {int t;      Read (t);  Init ();         while (t--) {int n,m;         Read (n); read (m);         int ans=0;                for (int i=min (40,m); i>=0;i--) {if (N>=f[i]) {ans+=n/f[i];            N=n%f[i];        }} printf ("%d\n", ans); } return 0;}

　　

hdu-5747 Aaronson (water problem)