目标
· 两种常用 I/O 模板:
- 数字流(StreamTokenizer) —— 适合大量数字、矩阵、直到 EOF 的多组数据
- 按行读取(BufferedReader + StringTokenizer) —— 适合“每行一个用例/表达式”的题
· 建议:
- 尽量用 BufferedReader / StreamTokenizer / PrintWriter,避免
Scanner和System.out(较慢) - 输出尽量用 StringBuilder + PrintWriter,只在最后一次性打印
- 需要频繁读写的大型结构,可用全局静态数组减少临时分配(避免反复 new),但仍计入内存限制(只是避免栈溢出和频繁 GC)
- 不推荐:临时动态空间
- 推荐:全局静态空间
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//全局静态空间,不会被卡内存限制
public static int MAXN = 201;
public static int MAXM = 201;
public static int[][] mat = new int[MAXN][MAXM];
public static int[] arr = new int[MAXM];
public static void main(String[] args) throws IOException{
//读入内存托管:把文件里的内容,load进来,保存在内存里,高效经济,托管的很好
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
//一个一个读数字
StreamTokenizer in = new StreamTokenizer(br);//不会区分空格和回车
//读出内存托管:提交答案的时候用,也是一个内存托管区
PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
while(in.nextToken() != StreamTokenizer.TT_EOF){//文件没有结束就继续
//n,二维数组的行
int n = (int) in.nval;
in.nextToken();
//m,二维数组的列
int m = (int) in.nval;
//装数字的矩阵,临时动态生成
int[][] mat = new int[n][m];
for(int i = 0; i<n;i++){
for(int j = 0; j<m ;j++){
in.nextToken();
mat[i][j]=(int)in.nval;//临时动态空间
}
}
out.println(maxSumSubmatrix(mat,n,m));
}
out.flush();
out.close();
}
public static void main(String[] args) throws IOException{
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
while((line = in.readLine()) != null){
parts = line.split(""); //根据空格切分
sum = 0; //求累加和
for(String num : parts){
sum+=Integer.valueOf(num); //把每一个数字都转成整形
}
out.println(sum);
}
out.flush();
in.close();
out.close();
}