44 lines
836 B
C++
44 lines
836 B
C++
#include <bits/stdc++.h>
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using namespace std;
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const int N = 11;
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int a[N] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
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void print() {
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int s = 0;
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for (int i = 1; i < N; i++) {
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s += a[i];
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cout << s % 2 << " ";
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}
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cout << "\n";
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}
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int main() {
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/*
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01矩阵+区间修改+单点查询 <=> 非01矩阵+单点修改+区间前缀和查询
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办法：
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① 区间[l,r]取反 => 单点修改：a[l]++,a[r+1]++
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② 单点查询a[k] => 区间前缀和查询： sum(a[1~k])%2
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*/
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//将2-4取反
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a[2]++, a[4 + 1]--;
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print();
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//将2-4再取一次反
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a[2]++, a[4 + 1]--;
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print();
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//将3-4再取一次反
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a[3]++, a[4 + 1]--;
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print();
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return 0;
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}
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/*
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输出：
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0 1 1 1 0 0 0 0 0 0
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0 0 0 0 0 0 0 0 0 0
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0 0 1 1 0 0 0 0 0 0
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*/ |