Created
January 17, 2014 17:02
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#include <cstdio> | |
using namespace std; | |
void parcel(int &box,int num[],int &space,int w); | |
int main() | |
{ | |
int num[7]; | |
while(scanf("%d %d %d %d %d %d",&num[1],&num[2],&num[3],&num[4],&num[5],&num[6])){ | |
int i=1; | |
for(;i<=6;i++) | |
if(num[i]!=0) break; | |
if(i==7) break; | |
int box=num[6]+num[5]+num[4]; | |
num[1]-=11*num[5]; //一個5x5搭配11個1x1 | |
num[2]-=5*num[4]; //一個4x4搭配5個2x2(如果2x2不夠的情況最底下會考慮) | |
box+=(num[3]/4); if(num[3]%4) box++; | |
switch(num[3]%4){ //3x3情況要特別討論 | |
case 0: break; | |
case 1: | |
num[2]-=5; | |
num[1]-=7; | |
break; | |
case 2: | |
num[2]-=3; | |
num[1]-=6; | |
break; | |
case 3: | |
num[2]-=1; | |
num[1]-=5; | |
break; | |
} | |
if (num[2]>0){ //如果2x2還有剩 | |
box+=num[2]/9; if(num[2]%9) box++; | |
num[1]-=(36-(num[2]%9)*4); | |
} | |
else if (num[2]<0) | |
num[1]-=(-num[2])*4; //將不夠的2x2用4個1x1來補 | |
if (num[1]>0){ //如果1x1還有剩 | |
box+=(num[1]/36); if(num[1]%36) box++; | |
} | |
printf("%d\n",box); | |
} | |
return 0; | |
} |
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