HDU 1850 Being a Good Boy in Spring Festival (Nim博弈求第一步选择数)
2016-08-05 20:34
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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1850
题意:Nim博弈,问先手的人如果想赢,第一步有几种选择。
必胜态下,a1^a2^.......^an=k,k不为零,根据定理:若a1^a2^...^an!=0,一定存在某个合法的移动,将ai改变成ai'后满足a1^a2^...^ai'^...^an=0。
分析一下,若a1^a2^...^an=k,则一定存在某个ai,它的二进制表示在k的最高位上是1,从而抑或后才能得到k。这时ai^k<ai一定成立。则我们可以将ai改变成ai'=ai^k,此时a1^a2^...^ai'^...^an=a1^a2^...^an^k=0。 因此,我们只需要看看有几个ai使得ai^k<ai成立即可。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int main() {
int m, n;
int i;
int a[110];
while(~scanf("%d", &m), m) {
int sum = 0;
for(i = 0; i < m; i++) {
scanf("%d", a + i);
sum ^= a[i];
}
int ans = 0;
if(sum == 0) puts("0");
else {
for(i = 0; i < m; i++) {
if((sum ^ a[i]) < a[i]) { // 注意^的优先级很低
ans++;
}
}
printf("%d\n", ans);
}
}
return 0;
}
题意:Nim博弈,问先手的人如果想赢,第一步有几种选择。
必胜态下,a1^a2^.......^an=k,k不为零,根据定理:若a1^a2^...^an!=0,一定存在某个合法的移动,将ai改变成ai'后满足a1^a2^...^ai'^...^an=0。
分析一下,若a1^a2^...^an=k,则一定存在某个ai,它的二进制表示在k的最高位上是1,从而抑或后才能得到k。这时ai^k<ai一定成立。则我们可以将ai改变成ai'=ai^k,此时a1^a2^...^ai'^...^an=a1^a2^...^an^k=0。 因此,我们只需要看看有几个ai使得ai^k<ai成立即可。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int main() {
int m, n;
int i;
int a[110];
while(~scanf("%d", &m), m) {
int sum = 0;
for(i = 0; i < m; i++) {
scanf("%d", a + i);
sum ^= a[i];
}
int ans = 0;
if(sum == 0) puts("0");
else {
for(i = 0; i < m; i++) {
if((sum ^ a[i]) < a[i]) { // 注意^的优先级很低
ans++;
}
}
printf("%d\n", ans);
}
}
return 0;
}
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