偶尔重抄旧业嘛
大意是给一个范围N，然后A和B会从1<=x<=N 中分别选出A1,A2,A3,B1,B2,B3；求有多少组3位数(C1,C2,C3)，满足A或者B中三位数一一对应位相差 <=2，注意数排列是一个环，像钟表走时一样。
可以直接遍历剪枝：
Version:1.0StartHTML:0000000128EndHTML:0000015312StartFragment:0000000128EndFragment:0000015312SourceURL:about:blank
/*
ID: wanglan1
LANG: JAVA
TASK: combo
*/
importjava.io.BufferedReader;
importjava.io.BufferedWriter;
importjava.io.FileReader;
importjava.io.FileWriter;
importjava.io.IOException;
importjava.io.PrintWriter;
public classcombo {
public static voidmain(String args[])throwsIOException {
BufferedReader f =newBufferedReader(newFileReader("combo.in"));
PrintWriter out =newPrintWriter(newBufferedWriter(newFileWriter("combo.out")));
intlen = Integer.valueOf(f.readLine());
String[] ssA = f.readLine().trim().split(" ");
String[] ssB = f.readLine().trim().split(" ");
f.close();
int[] a = {Integer.valueOf(ssA[0]),Integer.valueOf(ssA[1]),Integer.valueOf(ssA[2])};
int[] b = {Integer.valueOf(ssB[0]),Integer.valueOf(ssB[1]),Integer.valueOf(ssB[2])};
intcount =0,maxA,maxB,minA,minB;
for(inti =1;i <= len;i++) {
booleanAA00 =false,BB00 =false;
BB00 =false;
maxA =max(i,a[0]);
minA =min(i,a[0]);
maxB =max(i,b[0]);
minB =min(i,b[0]);
if(isCan(maxA,minA,len))
AA00 =true;
if(isCan(maxB,minB,len))
BB00 =true;
if(AA00 || BB00)
for(intj =1;j <= len;j++) {
booleanAA11 =false,BB11 =false;
maxA =max(j,a[1]);
minA =min(j,a[1]);
maxB =max(j,b[1]);
minB =min(j,b[1]);
if(AA00 &&isCan(maxA,minA,len))
AA11 =true;
if(BB00 &&isCan(maxB,minB,len))
BB11 =true;
if(AA11 || BB11)
for(intk =1;k <= len;k++) {
maxA =max(k,a[2]);
minA =min(k,a[2]);
maxB =max(k,b[2]);
minB =min(k,b[2]);
if(AA00 && AA11 &&isCan(maxA,minA,len) || BB00 && BB11 &&isCan(maxB,minB,len)) {
count++;
}
}
}
}
out.println(count);
out.close();
}
private static intmax(inta, intb) {
returna > b ? a : b;
}
private static intmin(inta, intb) {
returna > b ? b : a;
}
private static booleanisCan(inta, intb, intlen) {
if(a -2<= b || (a +2- len >= b))
return true;
return false;
}
}
