Codeforces 734F Anton and School
2016-11-22 18:06
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先由 x|y + x&y = x+y得到a序列。
因为b序列是&运算,用位运算可以用a序列反推回b序列进行验证。
代码:
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 2e5+5;
typedef __int64 ll;
ll a[MAXN], b[MAXN], c[MAXN], num[MAXN];
int main() {
int n;
cin >> n;
ll sum = 0;
bool flag = 1;
for(int i = 0; i < n; i++) scanf("%I64d", &b[i]), sum += b[i];
for(int i = 0; i < n; i++) scanf("%I64d", &c[i]), sum += c[i];
if(sum%(2*n) == 0) {
sum /= 2*n;
for(int i = 0; i < n; i++) {
ll t = b[i]+c[i];
if((t-sum)%n != 0) flag = 0;
a[i] = (t-sum)/n;
}
}
else flag = 0;
memset(num, 0, sizeof(num));
if(flag) {
for(int i = 0; i < n; i++) {
for(int j = 0; (1<<j) <= a[i]; j++) {
if(a[i] & (1<<j)) num[j]++;
}
}
for(int i = 0; i < n; i++) {
ll t = 0;
for(int j = 0; (1<<j) <= a[i]; j++) {
if(a[i] & (1<<j)) t += num[j]*(1LL<<j);
}
if(t != b[i]) flag = 0;
}
}
if(flag) {
for(int i = 0; i < n; i++) printf("%I64d ", a[i]);
}
else puts("-1");
return 0;
}
因为b序列是&运算,用位运算可以用a序列反推回b序列进行验证。
代码:
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 2e5+5;
typedef __int64 ll;
ll a[MAXN], b[MAXN], c[MAXN], num[MAXN];
int main() {
int n;
cin >> n;
ll sum = 0;
bool flag = 1;
for(int i = 0; i < n; i++) scanf("%I64d", &b[i]), sum += b[i];
for(int i = 0; i < n; i++) scanf("%I64d", &c[i]), sum += c[i];
if(sum%(2*n) == 0) {
sum /= 2*n;
for(int i = 0; i < n; i++) {
ll t = b[i]+c[i];
if((t-sum)%n != 0) flag = 0;
a[i] = (t-sum)/n;
}
}
else flag = 0;
memset(num, 0, sizeof(num));
if(flag) {
for(int i = 0; i < n; i++) {
for(int j = 0; (1<<j) <= a[i]; j++) {
if(a[i] & (1<<j)) num[j]++;
}
}
for(int i = 0; i < n; i++) {
ll t = 0;
for(int j = 0; (1<<j) <= a[i]; j++) {
if(a[i] & (1<<j)) t += num[j]*(1LL<<j);
}
if(t != b[i]) flag = 0;
}
}
if(flag) {
for(int i = 0; i < n; i++) printf("%I64d ", a[i]);
}
else puts("-1");
return 0;
}
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