长时间没码代码后，对题目明显变得生疏，还是需要找几道水题来练练手

1129: An Old Problem

思路：典型的矩阵交换行列的题，建立索引，利用交换索引替代直接对矩阵进行交换

#include <bits/stdc++.h>

using namespace std;

const int maxn = 1e3+10;

int a[maxn][maxn];

int x[maxn],y[maxn];

int main()
{
    int m,n,k;
    while(scanf("%d%d%d",&n,&m,&k)!=EOF)
    {
        for(int i=0;i<n;i++)
        {
            y[i] = i;
            for(int j=0;j<m;j++)
            {
                scanf("%d",&a[i][j]);
            }
        }
        for(int j=0;j<m;j++)    x[j] = j;
        while(k--)
        {
            int p,a,b;
            scanf("%d%d%d",&p,&a,&b);
            if(p==0)
                swap(y[a-1],y[b-1]);
            else
                swap(x[a-1],x[b-1]);
        }
        for(int i=0;i<n;i++)
        {
            printf("%d",a[y[i]][x[0]]);
            for(int j=1;j<m;j++)
            {
                printf(" %d",a[y[i]][x[j]]);
            }
            printf("\n");
            //printf("%d\n",a[y[i]][x[m-1]]);
        }
    }
    return 0;
}

1133: Special Judge

思路：就是个简单的模拟题，模拟整个判定过程即可

#include <bits/stdc++.h>

using namespace std;

const int maxn = 1e3+10;

typedef long long ll;

ll a[maxn],b[maxn];

int main()
{
    int k,al,bl;
    while(scanf("%d%d%d",&al,&bl,&k)!=EOF)
    {
        for(int i=0;i<al;i++)   scanf("%lld",&a[i]);
        for(int i=0;i<bl;i++)   scanf("%lld",&b[i]);
        int sit = -1;
        for(int i=2;i<bl;i++)
            if(b[i]-b[i-1]!=b[i-2])
                sit=0;
        if(sit==0)  printf("Wrong Answer - The participant's answer does not obey the recursion\n");
        else
        {
            int pos = 0;
            for(int i=0;i<al;i++)
            {
                if(a[i]==b[pos])
                    pos++;
                if(pos==bl)
                    break;
            }
            if(pos<bl)  printf("Wrong Answer - The participant's answer is not a subsequence of input\n");
            else if(bl>k)   printf("System Error - The participant gets better answer than jury's\n");
            else if(bl==k)   printf("Accepted - Ok, ok\n");
            else    printf("Wrong Answer - The jury gets better answer than participant's\n");
        }
    }
    return 0;
}

1292: World Game

思路：简单匹配题，从母串中匹配子序列，只需要统计字母次数即可，最终输出最长子序列

#include <bits/stdc++.h>

using namespace std;

const int maxn = 30;

char dic[maxn];
char alphabet[maxn];
int num[maxn];

int main()
{
    while(scanf("%s",dic)!=EOF)
    {
        int diclen = strlen(dic);
        int n;
        scanf("%d",&n);
        int re = 0;
        while(n--)
        {
            memset(num,0,sizeof num);
            for(int i=0;i<diclen;i++)    num[dic[i]-'a']++;
            scanf("%s",alphabet);
            int alplen = strlen(alphabet);
            int len = 0;
            for(int i=0;i<alplen;i++)
            {
                num[alphabet[i]-'a']--;
                if(num[alphabet[i]-'a']<0)  len = -1;
            }
            if(len==0&&alplen>re)   re = alplen;
        }
        printf("%d\n",re);
    }
    return 0;
}

1293: Arch0n's interesting game

思路：对牌取最小，找出第k大方案，一看就是个无法暴力的找规律题，那么就开始推公式咯

n*(n+1)/2=k
n^2+n=2*k
(n+1/2)^2=2*k+1/4
n=sqrt(2*k+1/4)-1/2

于是直接套公式即可，注意从大到小排列，向上取整，由于取最小最后得到的值+1即可

#include <bits/stdc++.h>

using namespace std;

const int maxn = 1e4+10;
const double eps = 1e-7;
int a[maxn];

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,k;
        scanf("%d%d",&n,&k);
        for(int i=0;i<n;i++)    scanf("%d",&a[i]);
        sort(a,a+n, greater<int>() );
        int re = (int)(sqrt(2.0*(double)(k)+1.0/4.0-eps) - 1.0/2.0)+1;
        printf("%d\n",a[re]);
    }
    return 0;
}