//这题很简单，字符串加模拟
//伪代码：
int n m t i j
ch a[][]
str st

    输入：n m st
    i <- j <- 1
    a[1][1] <- st[0];
    t<n*m-1 then do //模拟法造螺旋方阵
        j+1<=m and not a[i][j+1] then do a[i][++j]<-st[++t];
        i+1<=n and not a[i+1][j] then do a[++i][j]<-st[++t];
        j-1>=1 and not a[i][j-1] then do a[i][--j]<-st[++t];
        not a[i-1][j] then do a[--i][j]<-st[++t];
    输出一行，不加空格