Glad You CameTime Limit: 10000/5000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1829 Accepted Submission(s): 789 Problem Description Steve has an integer array a of length n (1-based). He assigned all the elements as zero at the beginning. After that, he made m operations, each of which is to update an interval of a with some value. You need to figure out ⨁ni=1(i⋅ai) after all his operations are finished, where ⨁ means the bitwise exclusive-OR operator.In order to avoid huge input data, these operations are encrypted through some particular approach. There are three unsigned 32-bit integers X,Y and Z which have initial values given by the input. A random number generator function is described as following, where ∧ means the bitwise exclusive-OR operator, << means the bitwise left shift operator and >> means the bitwise right shift operator. Note that function would change the values of X,Y and Z after calling.
Let the i -th result value of calling the above function as fi (i=1,2,⋯,3m) . The i -th operation of Steve is to update aj as vi if aj<vi (j=li,li+1,⋯,ri) , where ⎧⎩⎨⎪⎪lirivi=min((f3i−2modn)+1,(f3i−1modn)+1)=max((f3i−2modn)+1,(f3i−1modn)+1)=f3imod230(i=1,2,⋯,m). Input The first line contains one integer T , indicating the number of test cases. Output For each test case, output the answer in one line. Sample Input 4 1 10 100 1000 10000 10 100 1000 10000 100000 100 1000 10000 100000 1000000 1000 10000 100000 1000000 10000000 Sample Output 1031463378 1446334207 351511856 47320301347 Hint In the first sample, a = [1031463378] after all the operations. In the second sample, a = [1036205629, 1064909195, 1044643689, 1062944339, 1062944339, 1062944339, 1062944339, 1057472915, 1057472915, 1030626924] after all the operations.Source |
#include <bits/stdc++.h>
using namespace std;
#define ui unsigned int
#define ll long long
const int mn = 1e5 + 10, mm = 5e6 + 10;
ui X, Y, Z;
ui f[3 * mm];
ui st[mn][20];
ui RNG()
{
X = X ^ (X << 11);
X = X ^ (X >> 4);
X = X ^ (X << 5);
X = X ^ (X >> 14);
ui W = X ^ (Y ^ Z);
X = Y;
Y = Z;
Z = W;
return Z;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("C:\\in.txt", "r", stdin);
#endif // ONLINE_JUDGE
int T;
scanf("%d", &T);
while (T--)
{
memset(st, 0, sizeof st);
ui n, m;
scanf("%u %u %u %u %u", &n, &m, &X, &Y, &Z);
for (ui i = 1; i <= 3 * m; i++)
f[i] = RNG();
for (ui i = 1; i <= m; i++)
{
ui l = min(f[3 * i - 2] % n + 1, f[3 * i - 1] % n + 1);
ui r = max(f[3 * i - 2] % n + 1, f[3 * i - 1] % n + 1);
ui v = f[3 * i] % (1 << 30);
ui temp = log2(r - l + 1);
/// 更新 l 为起点 长度为 (1 << temp) 的区间
/// 更新 另一半长度为 (1 << temp) 的区间
st[l][temp] = max(v, st[l][temp]);
st[r - (1 << temp) + 1][temp] = max(v, st[r - (1 << temp) + 1][temp]);
}
/// 反向st表
for (ui i = log2(n); i >= 1; i--) /// 区间长度从大到小枚举更新
{
for (ui j = 1; j <= n; j++) /// 从前往后枚举
{
if ((j + (1 << i)) - 1 > n) /// 越界
break;
/// 更新 st[j][i] 的两半子集
st[j][i - 1] = max(st[j][i - 1], st[j][i]);
st[j + (1 << (i - 1))][i - 1] = max(st[j + (1 << (i - 1))][i - 1], st[j][i]);
}
}
ll sum = 0;
for (ui i = 1; i <= n; i++)
sum ^= (ll)i * (ll)st[i][0];
printf("%lld\n", sum);
}
return 0;
}