#include<iostream>
using namespace std;
const int N = 1e3 + 4;
int n, m, q;
int a[N][N], b[N][N];
void insert(int x1, int y1, int x2, int y2, int c){
//差分时以x2, y2 为扩展对象
b[x1][y1] += c;
b[x2 + 1][y...
| English Article | Word Count | Duration | Accuracy | Speed | Backspace | Error | Slogan |
|---|---|---|---|---|---|---|---|
| 《差分矩阵》 | 872字 | 12.0分钟 | 100% | 65 CPM | 126 次 | 1 | 不及格，加强练习！ |
| The wrong word this time: j | |||||||
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Suggestion:
可能对按键不够熟悉，或者对输入法不够熟练所导致，不要急慢慢来，先打对把准确率提高，然后逐渐的加快速度，最后才能打准打快。
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| Inspirational quotes: 领悟音乐的人，能从一切世俗的烦恼中超脱出来。 ——贝多芬 | |||||||
| What do you think of this article? Give it a rating and let it serve as a guiding light for others | |||||||
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