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留念,,,#include<cstdio> #include<algorithm> using namespace std; int C[7]={0}; //N[1]--N[6]分别代表1-6号物品的个数 int solve(){ int tol=C[6]; tol+=C[5]; C[1]=max(0,C[1]-11*C[5]); //1号尽量多的往5号放置 tol+=C[4]; if(C[2]>=C[4]*5) //2号箱子多 C[2]-=C[4]*5; else{ //2号箱子不够,用1号箱子补充 C[1]=max(0,C[1]-4*(5*C[4]-C[2])); C[2]=0; } tol+=(C[3]+3)/4; //每4个3号箱子放一个盒子,+3是为了防止小于4的时候,出现除法结果是0的情况 int mod=C[3]%4; C[3]=0; if(mod){ //3号由剩余 if(C[2]>=7-2*mod){ //由剩余规律得出的,先找数的规律,然后再进行计算 C[2]-=7-2*mod; C[1]=max(0,C[1]-(8-mod)); } else{ C[1]=max(0,C[1]-(36-9*mod-4*C[2])); //必然的,去掉3*3和2*2占用的,剩下的只能由1*1填补 C[2]=0; } } tol+=(C[2]+8)/9; //与C[3]+3的效果一样 mod=C[2]%9; C[2]=0; if(mod) C[1]=max(0,C[1]-(36-4*mod)); tol+=(C[1]+35)/36; return tol; } int main() { while(1){ int t=0; for(int i=1;i<=6;i++){ scanf("%d",&C[i]); t+=C[i]; } if(t==0)break; printf("%d\n",solve()); } return 0; } Followed by:
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