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  • POJ 3187 Backward Digit Sums 枚举

    用next_permutation枚举全排列。

    //#pragma comment(linker, "/STACK:1024000000,1024000000")
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<algorithm>
    #include<iostream>
    #include<sstream>
    #include<cmath>
    #include<climits>
    #include<string>
    #include<map>
    #include<queue>
    #include<vector>
    #include<stack>
    #include<set>
    using namespace std;
    typedef long long ll;
    typedef unsigned long long ull;
    typedef pair<int,int> pii;
    #define pb(a) push(a)
    #define INF 0x1f1f1f1f
    #define lson idx<<1,l,mid
    #define rson idx<<1|1,mid+1,r
    #define PI  3.1415926535898
    template<class T> T min(const T& a,const T& b,const T& c) {
        return min(min(a,b),min(a,c));
    }
    template<class T> T max(const T& a,const T& b,const T& c) {
        return max(max(a,b),max(a,c));
    }
    void debug() {
    #ifdef ONLINE_JUDGE
    #else
    
        freopen("in.txt","r",stdin);
        //freopen("d:\out1.txt","w",stdout);
    #endif
    }
    int getch() {
        int ch;
        while((ch=getchar())!=EOF) {
            if(ch!=' '&&ch!='
    ')return ch;
        }
        return EOF;
    }
    
    const int maxn=12;
    int ans[maxn];
    int n;
    int check()
    {
        int res[maxn][maxn];
        for(int i=0;i<n;i++)
            res[n][i]=ans[i];
        for(int i=n-1;i>=1;i--)
        {
            for(int j=0;j<i;j++)
                res[i][j]=res[i+1][j]+res[i+1][j+1];
        }
        return res[1][0];
    }
    int main()
    {
        int K;
        while(scanf("%d%d",&n,&K)!=EOF)
        {
            for(int i=1;i<=n;i++)
                ans[i-1]=i;
            do
            {
                if(check()==K)break;
            }while(next_permutation(ans,ans+n));
            for(int i=0;i<n;i++)
                printf("%d%c",ans[i],i==n-1?'
    ':' ');
        }
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/BMan/p/3648359.html
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