高斯消元法HDU - 5833 Zhu and 772002
【摘要】
目录
一，高斯消元法解异或方程
二，OJ实战
HDU - 5833 Zhu and 772002
UVA - 11542 Square
一，高斯消元法解异或方程
从左往右依次找到每一列的一个不为0的元素（如果有的话），用消元法消去这一列其他行的不为0的数，直到整个方程求解出来。
二，OJ实战
HDU...
目录
一，高斯消元法解异或方程
从左往右依次找到每一列的一个不为0的元素（如果有的话），用消元法消去这一列其他行的不为0的数，直到整个方程求解出来。
二，OJ实战
HDU - 5833 Zhu and 772002
题目：
Description
Zhu and 772002 are both good at math. One day, Zhu wants to test the ability of 772002, so he asks 772002 to solve a math problem.
But 772002 has a appointment with his girl friend. So 772002 gives this problem to you.
There are numbers . The value of the prime factors of each number does not exceed , you can choose at least one number and multiply them, then you can get a number .
How many different ways of choices can make is a perfect square number. The answer maybe too large, so you should output the answer modulo by 1000000007
Input
First line is a positive integer T, represents there are T test cases.
For each test case:
First line includes a number ，next line there are numbers .
Output
For the i-th test case , first output Case #i: in a single line.
Then output the answer of i-th test case modulo by 1000000007
Sample Input
2
3
3 3 4
3
2 2 2
Sample Output
Case #1:
3
Case #2:
3
我的思路是，把2000以内的303个素数用数组存起来，然后把输入的n个数表示成n行，每一行都是303个0或者1。
1表示的是这个数中这个素因子的次数为奇数，0表示为偶数。
现在就是要求，有多少种选行的方法，使得行的和全为0。
这里的和，可以理解为普通的和mod2，也可以理解为异或。
然后怎么求呢，这就是高斯消元法了。
本来应该能对的，可惜我傻了，偷懒了，没有真的把每个数表示成1个数组，用了差不多的形式。
然后就悲催了，那么多素数，它们的积一下就超了long long了，然后就wrong answer了。
错误代码：
#include <iostream>
#include <stdio.h>
#include<vector>
using namespace std;
long long v[305];
int prime[] = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, 1187, 1193, 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259, 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, 1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499, 1511, 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583, 1597, 1601, 1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, 1663, 1667, 1669, 1693, 1697, 1699, 1709, 1721, 1723, 1733, 1741, 1747, 1753, 1759, 1777, 1783, 1787, 1789, 1801, 1811, 1823, 1831, 1847, 1861, 1867, 1871, 1873, 1877, 1879, 1889, 1901, 1907, 1913, 1931, 1933, 1949, 1951, 1973, 1979, 1987, 1993, 1997, 1999 };
int p;
long long huajian(long long a)
{
for (int i = 0; i < 303; i++)
{
int pp = prime[i];
while (a % (pp*pp) == 0)a /= (pp*pp);
}
return a;
}
int main()
{
int t, n, r, num;
long long a;
cin >> t;
for (int i = 1; i <= t; i++)
{
cin >> n;
r = 1;
num = 0;
while (n--)
{
cin >> a;
a = huajian(a);
if (a == 1)r = r * 2 % 1000000007;
else v[num++] = a;
}
for (int i = 0; i < 303; i++)
{
if (num == 0)break;
p = prime[i];
int key;
for (key = 0; key < num; key++)if (v[key] % p == 0)break;
if (key<num)
{
for (int j = key + 1; j < num;j++)
{
if (v[j] % p == 0)v[j] = huajian(v[j] * v[key]);/超long long
if (v[j] == 1)
{
r = r * 2 % 1000000007;
v[j--] = v[num-- - 1];
}
}
v[key] = v[num-- - 1];
}
}
printf("Case #%d:\n%d\n", i, (r-1) % 1000000007);
}
return 0;
}
然后现在又按照二维数组来写，写完就AC了。
代码：
#include <iostream>
using namespace std;
int prime[303] = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, 1187, 1193, 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259, 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, 1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499, 1511, 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583, 1597, 1601, 1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, 1663, 1667, 1669, 1693, 1697, 1699, 1709, 1721, 1723, 1733, 1741, 1747, 1753, 1759, 1777, 1783, 1787, 1789, 1801, 1811, 1823, 1831, 1847, 1861, 1867, 1871, 1873, 1877, 1879, 1889, 1901, 1907, 1913, 1931, 1933, 1949, 1951, 1973, 1979, 1987, 1993, 1997, 1999 };
int x[300][303];
bool flag[300];
int t, n, p;
long long a;
int main()
{
cin >> t;
for (int i = 1; i <= t; i++)
{
cin >> n;
int row = 0;
while (n--)
{
cin >> a;
for (int i = 0; i < 303; i++)
{
p = prime[i];
while (a % (p*p) == 0)a /= (p*p);
x[row][i] = (a%p == 0);
}
flag[row] = true;
row++;
}
for (int i = 0; i < 303; i++)for (int j = 0; j < row; j++)if (flag[j] && x[j][i])
{
for (int k = j + 1; k < row; k++)if (flag[k] && x[k][i])
for (int xi = 0; xi < 303; xi++)x[k][xi] ^= x[j][xi];
flag[j] = false;
break;
}
int r = 1, ji;
for (int i = 0; i < row; i++)
{
ji = 0;
if (flag[i])ji = 1;
for (int j = 0; j < row; j++)if (x[i][j])ji = 0;
if (ji)r = r * 2 % 1000000007;
}
printf("Case #%d:\n%d\n", i, (r - 1) % 1000000007);
}
return 0;
}
UVA - 11542 Square
题目：
这个题目完全就是HDU - 5833 Zhu and 772002 的简化版
所以我直接把那个题目的代码改了一丢丢就可以AC了。
代码：
#include <iostream>
using namespace std;
int prime[303] = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, 1187, 1193, 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259, 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, 1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499, 1511, 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583, 1597, 1601, 1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, 1663, 1667, 1669, 1693, 1697, 1699, 1709, 1721, 1723, 1733, 1741, 1747, 1753, 1759, 1777, 1783, 1787, 1789, 1801, 1811, 1823, 1831, 1847, 1861, 1867, 1871, 1873, 1877, 1879, 1889, 1901, 1907, 1913, 1931, 1933, 1949, 1951, 1973, 1979, 1987, 1993, 1997, 1999 };
int x[300][303];
bool flag[300];
int t, n, p;
long long a;
int main()
{
cin >> t;
for (int i = 1; i <= t; i++)
{
cin >> n;
int row = 0;
while (n--)
{
cin >> a;
for (int i = 0; i < 303; i++)
{
p = prime[i];
while (a % (p*p) == 0)a /= (p*p);
x[row][i] = (a%p == 0);
}
flag[row] = true;
row++;
}
for (int i = 0; i < 303; i++)for (int j = 0; j < row; j++)if (flag[j] && x[j][i])
{
for (int k = j + 1; k < row; k++)if (flag[k] && x[k][i])
for (int xi = 0; xi < 303; xi++)x[k][xi] ^= x[j][xi];
flag[j] = false;
break;
}
long long r = 1, ji;
for (int i = 0; i < row; i++)
{
ji = 0;
if (flag[i])ji = 1;
for (int j = 0; j < row; j++)if (x[i][j])ji = 0;
if (ji)r = r * 2 ;
}
cout<<r-1<<endl;
}
return 0;
}
文章来源: blog.csdn.net，作者：csuzhucong，版权归原作者所有，如需转载，请联系作者。
原文链接：blog.csdn.net/nameofcsdn/article/details/115530383
【版权声明】本文为华为云社区用户转载文章，如果您发现本社区中有涉嫌抄袭的内容，欢迎发送邮件进行举报，并提供相关证据，一经查实，本社区将立刻删除涉嫌侵权内容，举报邮箱：
cloudbbs@huaweicloud.com
- 点赞
- 收藏
- 关注作者
评论（0）