c++洛谷P2089烤鸡

c++洛谷P2089烤鸡

P2089烤鸡
思路：暴力枚举即可，用10个for 刚开始没想到，想太复杂了，然后输出要递增，其实用for从头开始枚举就已经可以实现递增了

#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{int n,a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,ans=0;cin>>n;//为了使代码效率更高，进行剪枝if(n<10||n>30){cout<<0;} for(a0=1;a0<=3;a0++){for(a1=1;a1<=3;a1++){for(a2=1;a2<=3;a2++){for(a3=1;a3<=3;a3++){for(a4=1;a4<=3;a4++){for(a5=1;a5<=3;a5++){for(a6=1;a6<=3;a6++){for(a7=1;a7<=3;a7++){for(a8=1;a8<=3;a8++){for(a9=1;a9<=3;a9++){if(a0+a1+a2+a3+a4+a5+a6+a7+a8+a9==n)ans++;}}}}}}}}}	}cout<<ans<<endl;
for(a0=1;a0<=3;a0++){for(a1=1;a1<=3;a1++){for(a2=1;a2<=3;a2++){for(a3=1;a3<=3;a3++){for(a4=1;a4<=3;a4++){for(a5=1;a5<=3;a5++){for(a6=1;a6<=3;a6++){for(a7=1;a7<=3;a7++){for(a8=1;a8<=3;a8++){for(a9=1;a9<=3;a9++){if(a0+a1+a2+a3+a4+a5+a6+a7+a8+a9==n){printf("%d %d %d %d %d %d %d %d %d %dn",a0,a1,a2,a3,a4,a5,a6,a7,a8,a9);}}}}}}}}}}	}
}