纪中20日c组模拟赛T1 2121. 简单游戏

T1 2121. 简单游戏

(File IO): input:easy.in output:easy.out

时间限制: 1000 ms  空间限制: 262144 KB  具体限制  

Goto ProblemSet

题目描述

      Charles和sunny在玩一个简单的游戏。若给出1~n的一个排列A，则将A1、A2相加，A2、A3相加……An-1、An相加，则得到一组n-1个元素的数列B；再将B1、B2相加，B2、B3相加，Bn-2、Bn-1相加，则得到一组n-2个元素的数列……如此往复，最终会得出一个数T。而Charles和sunny玩的游戏便是，Charles给出n和T，sunny在尽可能短的时间内，找到能通过上述操作得到T且字典序最小的1~n的排列。（sunny大声说：“What  an easy game！”，接着几下就给出了解），Charles觉得没意思，就想和你玩，当然，你可以用一种叫做“电子计算机”的东西帮你。

输入

       本题有多组数据，对于每组数据：一行两个整数n（0<n<=20），t即最后求出来的数。两个0表示输入结束

输出

      对于每组测试数据输出一行n个整数，用空格分开，行尾无多余空格，表示求出来的满足要求的1~n的一个排列。

样例输入

4 16
3 9
0 0 

样例输出

3 1 2 4
1 3 2


对样例解释：
开始排列：  3     1      2      4
第一次操作：3+1=4  1+2=3  2+4=6
     得到：  4      3      6
第二次得到：     7     9
最后就是：        16

数据范围限制

数据保证有解。请注意加粗字体！
对于30%的数据，保证该组里的每个N都不超过10。
对于100%的数据，保证有每个N不超过20，且每组数据的个数不超过10。

Solution

首先可以发现,对于一个长度为n的排列,经过了n-1次邻项相加后,第i项被加了C(n-1,i)次(二项式定理).

所以可以先预处理杨辉三角形。

Algorithm1

使用next_permutation方便的计算地排列

但是没有剪枝...很慢的

Code1

简单的垃圾代码

#pragma GCC optimize(2)
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#define IL inline
using namespace std;
int tria[30][30]={
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,2,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,3,3,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,4,6,4,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,5,10,10,5,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,6,15,20,15,6,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,7,21,35,35,21,7,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,8,28,56,70,56,28,8,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,9,36,84,126,126,84,36,9,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,10,45,120,210,252,210,120,45,10,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,11,55,165,330,462,462,330,165,55,11,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,12,66,220,495,792,924,792,495,220,66,12,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,13,78,286,715,1287,1716,1716,1287,715,286,78,13,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,14,91,364,1001,2002,3003,3432,3003,2002,1001,364,91,14,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,15,105,455,1365,3003,5005,6435,6435,5005,3003,1365,455,105,15,1,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,16,120,560,1820,4368,8008,11440,12870,11440,8008,4368,1820,560,120,16,1,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,17,136,680,2380,6188,12376,19448,24310,24310,19448,12376,6188,2380,680,136,17,1,0,0,0,0,0,0,0,0,0,0,0,
0,1,18,153,816,3060,8568,18564,31824,43758,48620,43758,31824,18564,8568,3060,816,153,18,1,0,0,0,0,0,0,0,0,0,0,
0,1,19,171,969,3876,11628,27132,50388,75582,92378,92378,75582,50388,27132,11628,3876,969,171,19,1,0,0,0,0,0,0,0,0,0,
0,1,20,190,1140,4845,15504,38760,77520,125970,167960,184756,167960,125970,77520,38760,15504,4845,1140,190,20,1,0,0,0,0,0,0,0,0,
0,1,21,210,1330,5985,20349,54264,116280,203490,293930,352716,352716,293930,203490,116280,54264,20349,5985,1330,210,21,1,0,0,0,0,0,0,0,
0,1,22,231,1540,7315,26334,74613,170544,319770,497420,646646,705432,646646,497420,319770,170544,74613,26334,7315,1540,231,22,1,0,0,0,0,0,0,
0,1,23,253,1771,8855,33649,100947,245157,490314,817190,1144066,1352078,1352078,1144066,817190,490314,245157,100947,33649,8855,1771,253,23,1,0,0,0,0,0,
0,1,24,276,2024,10626,42504,134596,346104,735471,1307504,1961256,2496144,2704156,2496144,1961256,1307504,735471,346104,134596,42504,10626,2024,276,24,1,0,0,0,0,
0,1,25,300,2300,12650,53130,177100,480700,1081575,2042975,3268760,4457400,5200300,5200300,4457400,3268760,2042975,1081575,480700,177100,53130,12650,2300,300,25,1,0,0,0,
0,1,26,325,2600,14950,65780,230230,657800,1562275,3124550,5311735,7726160,9657700,10400600,9657700,7726160,5311735,3124550,1562275,657800,230230,65780,14950,2600,325,26,1,0,0,
0,1,27,351,2925,17550,80730,296010,888030,2220075,4686825,8436285,13037895,17383860,20058300,20058300,17383860,13037895,8436285,4686825,2220075,888030,296010,80730,17550,2925,351,27,1,0,
0,1,28,378,3276,20475,98280,376740,1184040,3108105,6906900,13123110,21474180,30421755,37442160,40116600,37442160,30421755,21474180,13123110,6906900,3108105,1184040,376740,98280,20475,3276,378,28,1
};
//triangle
int ans;
int arr[50],t,n;
IL int read()
{
    char ch;int x=0;
    ch=getchar();
    while(ch<'0'||ch>'9') ch=getchar();
    while(ch>='0'&&ch<='9') x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
    return x; 
}
int main()
{
    freopen("easy.in","r",stdin);
    freopen("easy.out","w",stdout);
    n=read();
    t=read();
    do{
        for(int i=1;i<=n;i++) 
            arr[i]=i;
        do{
            ans=0;
            int i;
            for(i=1;i<=n&&ans<=t;i++)
                ans+=arr[i]*tria[n][i];
            if(ans==t&&i>n){
                for(int j=1;j<=n;j++){
                    printf("%d",arr[j]);
                    if(j<n) printf(" "); 
                }
                cout<<endl;
                break;
            }
        }while(next_permutation(arr+1,arr+n+1));
        n=read();
        t=read();
    }while(n!=0||t!=0);
     return 0;
}

Algorithm2

使用dfs，对每一位进行判断，同时标记use（是否使用过此数）

这比next_permutation好在可以尽情剪枝——只要你能想到

Code2

这是剪枝1：如果当前的和（前depth个数的和）已经超过了t，就跳出。

也可以把

if(sum>t) return;

放到for循环里（这不是废话吗）

可以减少一点分支

按照题解上的说法，这样子只有40分（果然……）

#pragma GCC optimize(2)
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#define IL inline
using namespace std;
int tria[21][30]={
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,2,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,3,3,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,4,6,4,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,5,10,10,5,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,6,15,20,15,6,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,7,21,35,35,21,7,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,8,28,56,70,56,28,8,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,9,36,84,126,126,84,36,9,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,10,45,120,210,252,210,120,45,10,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,11,55,165,330,462,462,330,165,55,11,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,12,66,220,495,792,924,792,495,220,66,12,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,13,78,286,715,1287,1716,1716,1287,715,286,78,13,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,14,91,364,1001,2002,3003,3432,3003,2002,1001,364,91,14,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,15,105,455,1365,3003,5005,6435,6435,5005,3003,1365,455,105,15,1,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,16,120,560,1820,4368,8008,11440,12870,11440,8008,4368,1820,560,120,16,1,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,17,136,680,2380,6188,12376,19448,24310,24310,19448,12376,6188,2380,680,136,17,1,0,0,0,0,0,0,0,0,0,0,0,
0,1,18,153,816,3060,8568,18564,31824,43758,48620,43758,31824,18564,8568,3060,816,153,18,1,0,0,0,0,0,0,0,0,0,0,
0,1,19,171,969,3876,11628,27132,50388,75582,92378,92378,75582,50388,27132,11628,3876,969,171,19,1,0,0,0,0,0,0,0,0,0,
};
int use[21];
int ans;
int arr[50],t,n;
IL int read()
{
    char ch;int x=0;
    ch=getchar();
    while(ch<'0'||ch>'9') ch=getchar();
    while(ch>='0'&&ch<='9') x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
    return x; 
}
bool succ=0;
IL void dfs(int depth,int sum)
{
    if(sum>t) return;
    if(succ) return;
    if(depth==n){
        if(sum==t)
        {
            succ=1;
            for(int j=1;j<=n;j++){
                printf("%d",arr[j]);
                if(j<n) printf(" "); 
            }
            printf("
");
        }
        return;
    }
    for(int i=1;i<=n;i++)
    {
        if(!use[i])
        {
            use[i]=1;
            arr[depth+1]=i;
            dfs(depth+1,sum+i*tria[n][depth+1]);
            arr[depth+1]=0;
            use[i]=0;
        }
    }
}
int main()
{
//    freopen("easy.in","r",stdin);
//    freopen("easy.out","w",stdout);
    n=read();
    t=read();
    do{
        succ=0;
        dfs(0,0);
        n=read();
        t=read();
    }while(n!=0||t!=0);
     return 0;
}

Algorithm3

这是剪枝2：

由于杨辉三角形有对称性

比如，枚举到 2 4 1 3 时，其sum（邻项合并后的结果）会与 2 1 4 3 相同

那么可以加入以下判断：

if(depth>(n/2) && i<arr[n-depth]) continue;

代码翻译

如果当前要枚举的数即将放的位置超过了总长度的一半，且即将枚举的数值i小于与它关于这个二项式对称的那一项，那就说明这两项如果调换，总值不会发生变化，那么就不考虑这种情况。

这句话真长……

Code3

#pragma GCC optimize(2)
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#define IL inline
using namespace std;
int tria[21][30]={
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,2,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,3,3,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,4,6,4,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,5,10,10,5,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,6,15,20,15,6,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,7,21,35,35,21,7,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,8,28,56,70,56,28,8,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,9,36,84,126,126,84,36,9,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,10,45,120,210,252,210,120,45,10,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,11,55,165,330,462,462,330,165,55,11,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,12,66,220,495,792,924,792,495,220,66,12,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,13,78,286,715,1287,1716,1716,1287,715,286,78,13,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,14,91,364,1001,2002,3003,3432,3003,2002,1001,364,91,14,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,15,105,455,1365,3003,5005,6435,6435,5005,3003,1365,455,105,15,1,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,16,120,560,1820,4368,8008,11440,12870,11440,8008,4368,1820,560,120,16,1,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,17,136,680,2380,6188,12376,19448,24310,24310,19448,12376,6188,2380,680,136,17,1,0,0,0,0,0,0,0,0,0,0,0,
0,1,18,153,816,3060,8568,18564,31824,43758,48620,43758,31824,18564,8568,3060,816,153,18,1,0,0,0,0,0,0,0,0,0,0,
0,1,19,171,969,3876,11628,27132,50388,75582,92378,92378,75582,50388,27132,11628,3876,969,171,19,1,0,0,0,0,0,0,0,0,0,
};
int use[21];
int arr[50],t,n;
IL int read()
{
    char ch;int x=0;
    ch=getchar();
    while(ch<'0'||ch>'9') ch=getchar();
    while(ch>='0'&&ch<='9') x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
    return x; 
}
bool succ=0;
IL void dfs(int depth,int sum)
{
    if(succ) return;
    if(depth==n){
        if(sum==t)
        {
            succ=1;
            for(int j=1;j<=n;j++){
                printf("%d",arr[j]);
                if(j<n) printf(" "); 
            }
            printf("
");
        }
        return;
    }
    for(int i=1;i<=n;i++)
    {
        if(!use[i])
        {
            if(depth>(n/2) && i<arr[n-depth]) continue;
            if(sum+i*tria[n][depth+1]>t) continue;
            use[i]=1;
            arr[depth+1]=i;
            dfs(depth+1,sum+i*tria[n][depth+1]);
            arr[depth+1]=0;
            use[i]=0;
        }
    }
}
int main()
{
//    freopen("easy.in","r",stdin);
//    freopen("easy.out","w",stdout);
    n=read();
    t=read();
    do{
        succ=0;
        dfs(0,0);
        n=read();
        t=read();
    }while(n!=0||t!=0);
     return 0;
}

好的，果然如题解所说，这个剪枝并不能加分。

Algorithm4

这是剪枝3：枚举前，就当前还未使用的数字来讲，可以将它们重新排序，对应还没使用的系数，可以算出已选择的数字不变的情况下，让之后几个数排列组合，再乘以对应系数的和的最大值与最小值。

貌似我讲的不太清楚。

当枚举到第X个数时，剩余的N-X个数，与剩余的N-X个系数一一对应，让大数和大系数相乘，小数和小系数相乘可以得到最大值，让大数和小系数相乘，小数和大系数相乘可以得到最小值，如果剩余的值不在这个范围内，就不要搜下去，这样可以大大优化。

举个栗子

若N=4，T=16：

当枚举arr[1]=1时，剩余16-1*1=15，剩余的未放置的数为2，3，4，剩余的系数为1，3，3，这样最大值为4*3+3*3+2*1=23，最小值为4*1+3*3+2*3=19，都超过了15，所以第一个数不能选1。

Code4

#pragma GCC optimize(2)
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#define IL inline
using namespace std;
int tria[21][30]={
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,2,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,3,3,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,4,6,4,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,5,10,10,5,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,6,15,20,15,6,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,7,21,35,35,21,7,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,8,28,56,70,56,28,8,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,9,36,84,126,126,84,36,9,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,10,45,120,210,252,210,120,45,10,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,11,55,165,330,462,462,330,165,55,11,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,12,66,220,495,792,924,792,495,220,66,12,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,13,78,286,715,1287,1716,1716,1287,715,286,78,13,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,14,91,364,1001,2002,3003,3432,3003,2002,1001,364,91,14,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,15,105,455,1365,3003,5005,6435,6435,5005,3003,1365,455,105,15,1,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,16,120,560,1820,4368,8008,11440,12870,11440,8008,4368,1820,560,120,16,1,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,17,136,680,2380,6188,12376,19448,24310,24310,19448,12376,6188,2380,680,136,17,1,0,0,0,0,0,0,0,0,0,0,0,
0,1,18,153,816,3060,8568,18564,31824,43758,48620,43758,31824,18564,8568,3060,816,153,18,1,0,0,0,0,0,0,0,0,0,0,
0,1,19,171,969,3876,11628,27132,50388,75582,92378,92378,75582,50388,27132,11628,3876,969,171,19,1,0,0,0,0,0,0,0,0,0,
};
int use[21];
int tarr[50],ttria[30];
int arr[50],t,n;
IL int read()
{
    char ch;int x=0;
    ch=getchar();
    while(ch<'0'||ch>'9') ch=getchar();
    while(ch>='0'&&ch<='9') x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
    return x; 
}
bool succ=0;
IL void dfs(int depth,int sum)
{
    if(succ) return;
    if(depth==n){
        if(sum==t)
        {
            succ=1;
            for(int j=1;j<=n;j++){
                printf("%d",arr[j]);
                if(j<n) printf(" "); 
            }
            printf("
");
        }
        return;
    }

    int s=0;
    for(int i=1;i<=n;i++)
    {
        if(use[i]) continue;
        tarr[++s]=i;
    }
    for(int i=depth+1;i<=n;i++)
        ttria[i-depth]=tria[n][i];
    sort(tarr+1,tarr+s+1);
    sort(ttria+1,ttria+s+1);
    int maxs=0,mins=0;
    for(int i=1;i<=s;i++)
    {
        maxs+=tarr[i]*ttria[i];
        mins+=tarr[i]*ttria[s-i+1];
    }
    if(t<mins+sum||t>maxs+sum) return;

    for(int i=1;i<=n;i++)
    {
        if(!use[i])
        {
            if(depth>(n/2) && i<arr[n-depth]) continue;
            if(sum+i*tria[n][depth+1]>t) continue;
            use[i]=1;
            arr[depth+1]=i;
            dfs(depth+1,sum+i*tria[n][depth+1]);
            arr[depth+1]=0;
            use[i]=0;
        }
    }
}
int main()
{
//    freopen("easy.in","r",stdin);
//    freopen("easy.out","w",stdout);
    n=read();
    t=read();
    do{
        succ=0;
        dfs(0,0);
        n=read();
        t=read();
    }while(n!=0||t!=0);
     return 0;
}

Impression

好辛苦呀~~~