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坑爹啊,中间输出没去掉,检查了好久……

略恶心的一道题- -

#include<stdio.h>
#include<string.h>
#include<math.h>
int main()
{
    int st[30][30],a,b,n,top[30],i,j;
    char s1[10],s2[10];
    while(scanf("%d",&n)!=EOF)
    {
        for(i=0;i<n;i++)
        {
            st[i][0]=i;
            top[i]=1;
        }
        while(scanf("%s",s1)&&strcmp(s1,"quit"))
        {
            scanf("%d%s%d",&a,s2,&b);
            if(strcmp(s1,"move")==0)
            {
                if(strcmp(s2,"onto")==0)
                {
                    for(i=0;i<n;i++)
                    {
                        for(j=0;j<top[i];j++)
                            if(st[i][j]==a)
                                break;
                        if(j!=top[i])
                            break;
                    }
                    int x=i,y=j;
                    for(i=0;i<n;i++)
                    {
                        for(j=0;j<top[i];j++)
                            if(st[i][j]==b)
                                break;
                        if(j!=top[i])
                            break;
                    }
                    if(x==i)
                        continue;
                    int y1=j;
                    for(j=y+1;j<top[x];j++)
                    {
                        st[st[x][j]][0]=st[x][j];
                        top[st[x][j]]=1;
                    }
                    top[x]=y;
                    for(j=y1+1;j<top[i];j++)
                    {
                        st[st[i][j]][0]=st[i][j];
                        top[st[i][j]]=1;
                    }
                    st[i][y1+1]=a;
                    top[i]=y1+2;
                }
                else if(strcmp(s2,"over")==0)
                {
                    for(i=0;i<n;i++)
                    {
                        for(j=0;j<top[i];j++)
                            if(st[i][j]==a)
                                break;
                        if(j!=top[i])
                            break;
                    }
                    int x=i,y=j;
                    for(i=0;i<n;i++)
                    {
                        for(j=0;j<top[i];j++)
                            if(st[i][j]==b)
                                break;
                        if(j!=top[i])
                            break;
                    }
                    if(x==i)
                        continue;
                    for(j=y+1;j<top[x];j++)
                    {
                        st[st[x][j]][0]=st[x][j];
                        top[st[x][j]]=1;
                    }
                    top[x]=y;
                    st[i][top[i]]=a;
                    top[i]++;
                }
            }
            else if(strcmp(s1,"pile")==0)
            {
                if(strcmp(s2,"onto")==0)
                {
                    for(i=0;i<n;i++)
                    {
                        for(j=0;j<top[i];j++)
                            if(st[i][j]==b)
                                break;
                        if(j!=top[i])
                            break;
                    }
                    int x=i,y=j;                   
                    for(i=0;i<n;i++)
                    {
                        for(j=0;j<top[i];j++)
                            if(st[i][j]==a)
                                break;
                        if(j!=top[i])
                            break;
                    }
                    int y1=j;
                    if(x==i)
                        continue;
                    for(j=y+1;j<top[x];j++)
                    {
                        st[st[x][j]][0]=st[x][j];
                        top[st[x][j]]=1;
                    }
                    top[x]=y+1;
                    for(j=y1;j<top[i];j++)
                        st[x][top[x]++]=st[i][j];
                    top[i]=y1;
                }
                else if(strcmp(s2,"over")==0)
                {
                    for(i=0;i<n;i++)
                    {
                        for(j=0;j<top[i];j++)
                            if(st[i][j]==b)
                                break;
                        if(j!=top[i])
                            break;
                    }
                    int x=i;
                    for(i=0;i<n;i++)
                    {
                        for(j=0;j<top[i];j++)
                            if(st[i][j]==a)
                                break;
                        if(j!=top[i])
                            break;
                    }
                    int y=j;
                    if(x==i)
                        continue;
                    for(;j<top[i];j++)
                        st[x][top[x]++]=st[i][j];
                    top[i]=y;
                    
                }
            } 
        }
        for(i=0;i<n;i++)
        {
            printf("%d:",i);
            for(j=0;j<top[i];j++)
                printf(" %d",st[i][j]);
            printf("\n");
        }
    }
    return 0;
}


别再折腾虚拟机了!手把手教你用WSL2在Windows 11上安装Kali Linux(含汉化与图形界面) 本文详细介绍了如何在Windows 11上通过WSL2安装和配置Kali Linux,包括汉化与图形界面设置。通过WSL2,用户可以享受接近原生性能的Linux体验,同时保持与Windows系统的无缝集成,特别适合网络安全爱好者和开发者。文章还提供了常见问题解决和性能优化建议,帮助用户高效使用Kali工具集。 阅读详情

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% Dropout fraction opts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samplesnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex4 neural net with sigmoid activation functionrand('state',0)nn = nnsetup([784 100 10]);nn.activation_function = 'sigm'; % Sigmoid activation functionnn.learningRate = 1; % Sigm require a lower learning rateopts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samplesnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex5 plotting functionalityrand('state',0)nn = nnsetup([784 20 10]);opts.numepochs = 5; % Number of full sweeps through datann.output = 'softmax'; % use softmax outputopts.batchsize = 1000; % Take a mean gradient step over this many samplesopts.plot = 1; % enable plottingnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex6 neural net with sigmoid activation and plotting of validation and training error% split training data into training and validation datavx = train_x(1:10000,:);tx = train_x(10001:end,:);vy = train_y(1:10000,:);ty = train_y(10001:end,:);rand('state',0)nn = nnsetup([784 20 10]); nn.output = 'softmax'; % use softmax outputopts.numepochs = 5; % Number of full sweeps through dataopts.batchsize = 1000; % Take a mean gradient step over this many samplesopts.plot = 1; % enable plottingnn = nntrain(nn, tx, ty, opts, vx, vy); % nntrain takes validation set as last two arguments (optionally)[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');```[![Bitdeli Badge](https://d2weczhvl823v0.cloudfront.net/rasmusbergpalm/deeplearntoolbox/trend.png)](https://bitdeli.com/free "Bitdeli Badge")

UVA101 The Blocks Problem 题解

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Uva 101 - The Blocks Problem

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arex71991的博客 131

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