題意は、1~value を構成するためにはそれぞれ一つの方法で支払う必要があるということです。
考えるべき問題で、方法を思いついたらすぐに通過できます。少し権重に似ています。
// BEGIN CUT HERE
// END CUT HERE
#line 5 "GCAcode.cpp"
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string\>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>
using namespace std;
#ifdef DEBUG
#define VAR(a,b) \_\_typeof(b) a=(b)
#define debug(...) printf("DEBUG: "),printf(\_\_VA\_ARGS\_\_)
#else
#define VAR(a,b) \_\_typeof(b) a=(b)
#define debug(...)
#endif
typedef unsigned int uint;
typedef long long int Int;
typedef unsigned long long int UInt;
#define Set(a,s) memset(a,s,sizeof(a))
#define Pln() printf("\\n")
#define For(i,x)for(int i=0;i<x;i++)
#define CON(x,y) x##y
#define M MAX\_LIMITS
#define PB push\_back
#define oo INT\_MAX
#define FOR(a,b) for(VAR(a,(b).begin());a!=(b).end();++a)
#define eps 1e-9
#define X first
#define Y second
inline bool xdy(double x,double y){return x>y+eps;}
inline bool xddy(double x,double y){return x>y-eps;}
inline bool xcy(double x,double y){return x<y-eps;}
inline bool xcdy(double x,double y){return x<y+eps;}
const Int mod=1000000007;
bool cmp(int x,int y){return x>y;}
class ChangePurse
{
public:
vector <int\> optimalCoins(vector <int\> coinTypes, int value){
vector<int\> ans;
map<int,int\> sid;
int n=coinTypes.size();
vector<int\> sortcoin=coinTypes;
sort(sortcoin.begin(),sortcoin.end(),cmp);
int now=value;
for(int i=0;i<n;i++){
if(now%sortcoin\[i\]==sortcoin\[i\]-1){
sid\[sortcoin\[i\]\]=now/sortcoin\[i\];
now=sortcoin\[i\]-1;
}
}
for(int i=0;i<n;i++){
ans.PB(sid\[coinTypes\[i\]\]);
}
// printf("%d\\n",n);
return ans;
}
// BEGIN CUT HERE
public:
void run\_test(int Case) {
if ((Case == -1) || (Case == 0)) test\_case\_0();
if ((Case == -1) || (Case == 1)) test\_case\_1();
if ((Case == -1) || (Case == 2)) test\_case\_2();
if ((Case == -1) || (Case == 3)) test\_case\_3();
if ((Case == -1) || (Case == 4)) test\_case\_4();
if ((Case == -1) || (Case == 5)) test\_case\_5();
if ((Case == -1) || (Case == 6)) test\_case\_6();
if ((Case == -1) || (Case == 7)) test\_case\_7();
}
private:
template <typename T> string print\_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const\_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\\"' << \*iter << "\\","; os << " }"; return os.str(); }
void verify\_case(int Case, const vector <int\> &Expected, const vector <int\> &Received) { cout << "Test Case #" << Case << "..."; if (Expected == Received) cout << "PASSED" << endl; else { cout << "FAILED" << endl; cout << "\\tExpected: " << print\_array(Expected) << endl; cout << "\\tReceived: " << print\_array(Received) << endl; } }
void test\_case\_0() {
int Arr0\[\] = {1,25,10};
vector <int\> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0\[0\])));
int Arg1 = 49;
int Arr2\[\] = { 24, 1, 0 };
vector <int\> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2\[0\])));
verify\_case(0, Arg2, optimalCoins(Arg0, Arg1)); }
void test\_case\_1() {
int Arr0\[\] = {1,7};
vector <int\> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0\[0\])));
int Arg1 = 49;
int Arr2\[\] = { 49, 0 };
vector <int\> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2\[0\])));
verify\_case(1, Arg2, optimalCoins(Arg0, Arg1)); }
void test\_case\_2() {
int Arr0\[\] = {11,5,10,1};
vector <int\> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0\[0\])));
int Arg1 = 109;
int Arr2\[\] = { 9, 0, 0, 10 };
vector <int\> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2\[0\])));
verify\_case(2, Arg2, optimalCoins(Arg0, Arg1)); }
void test\_case\_3() {
int Arr0\[\] = {29210, 58420, 350520, 708072, 720035, 230, 42355,
1, 59006, 985, 236024, 163, 701040};
vector <int\> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0\[0\])));
int Arg1 = 929579039;
int Arr2\[\] = { 1, 5, 1, 0, 0, 126, 0, 229, 0, 0, 0, 0, 1325 };
vector <int\> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2\[0\])));
verify\_case(3, Arg2, optimalCoins(Arg0, Arg1)); }
void test\_case\_4() {
int Arr0\[\] = {795, 5536, 26, 915, 18590, 60840, 49140, 2,
119700, 162235, 369000, 383936, 478800, 505995,
949, 95984, 455, 8, 420, 239400, 276800, 191968,
619305, 654810, 706420, 393120, 738000, 767872,
425880, 786240, 830400, 676, 4500, 851760, 957600,
648940, 1, 112, 180, 457};
vector <int\> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0\[0\])));
int Arg1 = 687245439;
int Arr2\[\] = { 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 1, 0, 0, 0, 1, 0, 13, 0, 0, 0, 1, 0, 0, 0, 0, 0, 894, 0, 0, 0, 0, 0, 0, 0, 0, 1, 856, 0, 0 };
vector <int\> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2\[0\])));
verify\_case(4, Arg2, optimalCoins(Arg0, Arg1)); }
void test\_case\_5() {
int Arr0\[\] = {494208, 722376, 731798, 809064, 920448, 1, 988416, 9152, 158,
991014, 282720, 40132, 608, 143, 289755, 734, 579510, 828400,
330338, 816, 460224, 27456, 675783, 331, 436, 82368, 729, 306,
202266, 247104, 414200, 705};
vector <int\> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0\[0\])));
int Arg1 = 419088383;
int Arr2\[\] = { 1, 0, 0, 0, 0, 142, 423, 2, 0, 0, 0, 0, 0, 63, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0 };
vector <int\> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2\[0\])));
verify\_case(5, Arg2, optimalCoins(Arg0, Arg1)); }
void test\_case\_6() {
int Arr0\[\] = {1,25,10};
vector <int\> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0\[0\])));
int Arg1 = 49;
int Arr2\[\] = { 24, 1, 0 };
vector <int\> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2\[0\])));
verify\_case(6, Arg2, optimalCoins(Arg0, Arg1)); }
void test\_case\_7() {
int Arr0\[\] = {1,25,10};
vector <int\> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0\[0\])));
int Arg1 = 49;
int Arr2\[\] = { 24, 1, 0 };
vector <int\> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2\[0\])));
verify\_case(7, Arg2, optimalCoins(Arg0, Arg1)); }
// END CUT HERE
};
// BEGIN CUT HERE
int main()
{
ChangePurse \_\_\_test;
\_\_\_test.run\_test(-1);
return 0;
}
// END CUT HERE