历届试题 高僧斗法

 //神奇的nim游戏....

 
#include<iostream>

#include<cstdio>

#include<string>

using namespace std;

int a[110];

int ge[110], ne[110];//ge为0-1,2-3  ne为1-2,3-4..... 

int main() {

	int n = 0;	//数字下标 

	while(~scanf("%d",&a[n]))	n++;

	

	int gn = 0;

	int sum = 0;
	

	for(int i = 1; i < n; i=i+2){

		ge[gn] = a[i]-a[i-1]-1;
		if(i+1 < n)
			ne[gn] = a[i+1] - a[i]-1;

		sum ^= ge[gn];

		gn++;

	}

	

	if(sum == 0) printf("-1");

	else{
		int ok = 0;

		for(int i = 0; i < n; i+=2){
/*
			for(int j = 1; j < a[i+1]-a[i]; j++){
//因为之前pm是按i+=2上来的  所以 若i为偶数，i/2为ge[]的左端点，否则，为ge[]右端点 
				if(i%2 == 0){
					if((sum^ge[i/2]^(ge[i/2]-j)) == 0){//左端点 往上走，距离减小 
						printf("%d %d",a[i],a[i]+j);
						ok = 1;
					}

				}else{

					if((sum^ge[i/2]^(ge[i/2]+j)) == 0){//右端点往上走，距离增加 //1^1 = 0即^原来的值，取消原来制的影响，引入新值 

						printf("%d %d",a[i],a[i]+j);

						ok = 1;

					}

				}
*/

int x =  sum^ge[i/2], l = ge[i/2], r = ne[i/2]+ge[i/2];
//cout << x << " l" << l << " r" << r << endl; 
					if(x  <= l){//左端点 往上走，距离减小 
						printf("%d %d", a[i], a[i]+l-x);
						ok = 1;
					}
					  else if(x <= r){//右端点往上走，距离增加 //1^1 = 0即^原来的值，取消原来制的影响，引入新值 

						printf("%d %d", a[i+1], a[i+1]-l+x);

						ok = 1;

					}			

			if(ok) break;

		}

	}

	return 0;

}