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P1147 连续自然数和

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https://www.luogu.com.cn/record/165820608
This commit is contained in:
Baoshuo Ren 2024-07-13 22:57:59 +08:00
parent df2ee95279
commit bcf4b5e34e
Signed by: baoshuo
GPG Key ID: 00CB9680AB29F51A
1 changed files with 14 additions and 9 deletions
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Luogu/P1147/P1147.cpp
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@ -2,22 +2,27 @@
using std::cin; using std::cin;
using std::cout; using std::cout;
#define endl '\n' const char endl = '\n';
int m, sum, j; int m;
int main() { int main() {
std::ios::sync_with_stdio(false); std::ios::sync_with_stdio(false);
cin.tie(nullptr);
cin >> m; cin >> m;
for (int i = 1; i <= m / 2; i++) {
sum = 0; // 为什么到 m/2 结束？因为 m/2 之后的数不可能连着取两个数之和等于 m。
for (j = i; j < m; j++) { for (int i = 1, j = 2, sum = 3 /* 初始 1 + 2 = 3 */; i <= (m >> 1);) {
sum += j; if (sum == m) { // 够了，输出答案，并且把左面的数从窗口中删去
if (sum >= m) break;
}
if (sum == m) {
cout << i << ' ' << j << endl; cout << i << ' ' << j << endl;
sum -= i++; // 压行：sum -= i; i++;
} else if (sum < m) { // 不够，需要往后再取一个数
sum += ++j; // 压行：j++; sum += j; # 注意：这里是先加 j 再加 sum，顺序要写对
} else { // sum > m # 多了，把左面的数从窗口中删去
sum -= i++; // 压行：sum -= i; i++;
} }
} }
return 0; return 0;
} }