由于之前此题java超时，最后曲线救国，刷了无数题之后，找到了一个java快速io

源码先贴出来 已ac

import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.StreamTokenizer; public class Main { static StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in))); static int nextInt() throws IOException{in.nextToken(); return (int) in.nval;} static String next() throws IOException{in.nextToken(); return in.sval;} public static void main(String[] args) throws IOException { int gradePb[], gradePs = 0, n, m, wrongCnt[]; int maxCnt = 0; String str, ansStr[]; StringBuilder sb = new StringBuilder(); char ch; n = nextInt(); m = nextInt(); gradePb = new int[m]; wrongCnt = new int[m]; ansStr = new String[m]; int a, b, c, ttemp; for(int i = 0; i < m; i++) { a = nextInt(); b = nextInt(); c = nextInt(); ttemp = c; sb.append(c); while(ttemp-- > 0) { str = next(); sb.append(str); } ansStr[i] = sb.toString(); gradePb[i] = a; sb.delete(0, sb.length()); } for(int i = 0; i < n; i++) { for(int j = 0; j < m; j++) { in.nextToken(); while((char)in.ttype != ')') { if(in.ttype == StreamTokenizer.TT_NUMBER) sb.append((int)in.nval); if(in.ttype == StreamTokenizer.TT_WORD) sb.append(in.sval); in.nextToken(); } str = sb.toString(); sb.delete(0, sb.length()); if(ansStr[j].equals(str)) gradePs+=gradePb[j]; else wrongCnt[j]++; } System.out.println(gradePs); gradePs = 0; } for(int i = 0; i < m; i++) { if(maxCnt < wrongCnt[i]) maxCnt = wrongCnt[i]; } if(maxCnt > 0) { System.out.print(maxCnt); for(int i = 1; i <= m; i++) { if(wrongCnt[i-1] == maxCnt) System.out.print(" "+i); } } else { System.out.print("Too simple"); } // System.out.println("all time is :"+(end-start)+"ms"); } }

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思悟

提交次数就不说了。。。简直要怀疑人生了。用尽各种javaIO的操作，优化再优化，无奈，java无法像c/c++可以非常便捷轻快的边读取数据边处理，但是不得不说javaIO的操作还是很舒服的，如果没有这么大时间消耗的话会更好吧！