uva 138 - Street Numbers(等差数列求和)
来源:互联网 发布:程序员源代码下载网站 编辑:程序博客网 时间:2024/04/28 14:36
题目链接:138 - Street Numbers
题目大意:找到10组a和b,使得sum[1,a - 1] == sum[a, b]。
解题思路:先用二分做了一遍,枚举b,二分a,但是这样很慢,这能求出答案后把表输出来。
二分:
#include <stdio.h>#include <string.h>long long count(long long a, long long b) {long long sum = (a + b) * (b - a + 1) / 2;return sum;}bool search(int b) {long long p, q;int l = 0, r = b;while (l < r) {int mid = (l + r) / 2;p = count(1, mid - 1), q = count(mid + 1, b);if (p < q) l = mid;else if (p > q) r = mid;else {printf("%10d%10d\n", mid, b);return true;}if (l == r - 1) break;}return false;}int main () {int cnt = 0;for (int i = 4; cnt < 10; i++)if (search(i)) cnt++;return 0;}
#include <stdio.h>int x[] = {6, 35, 204, 1189, 6930, 40391, 235416, 1372105, 7997214, 46611179};int y[] = {8, 49, 288, 1681, 9800, 57121, 332928, 1940449, 11309768, 65918161};int main () {for (int i = 0; i < 10; i++)printf("%10d%10d\n", x[i], y[i]); return 0;}
现在有个数学公式,根据等差数列求和公式得到的,假设存在m和n满足上述条件,那么sum(1, m - 1 = (m - 1) * m / 2,
sum(m + 1, n) = (m + 1 + n) * (n - m) / 2, 两边相等化简得到n = ( sqrt(8 * m * m + 1) - 1) / 2.
#include <stdio.h>#include <math.h>int main() {long long m = 1,c = 0;double N;while(c < 10) {m++;N = (sqrt(1.0 + 8 * m * m) - 1)/2;if(N == floor(N)){printf("%10lld",m);printf("%10.0lf\n",N);c++;}}return 0;}
- uva 138 - Street Numbers(等差数列求和)
- uva 138 - Street Numbers
- uva 138 street numbers
- UVA 138 Street Numbers
- UVA 138 Street Numbers
- UVA 138 Street Numbers
- UVa 138 - Street Numbers
- UVa 138 - Street Numbers
- UVa 138: Street Numbers
- UVA - 138 Street Numbers
- UVA 138 - Street Numbers
- UVa 138 Street Numbers
- uva 138 Street Numbers
- UVA - 138 Street Numbers
- UVA 138 Street Numbers
- UVA138 - Street Numbers(等差数列)
- UVA 138 - Street Numbers (数论)
- UVA - 138 - Street Numbers (简单数论)
- 【博弈学习】
- HDU 3874 Necklace 树状数组 + 离线处理
- Android Activity 常用功能设置(全屏、横竖屏等)
- 数据库的集中备份模式含义
- Android Camera开发:给摄像头预览界面加个ZoomBar(附完整代码下载)
- uva 138 - Street Numbers(等差数列求和)
- Log.v Log.d Log.e Log.i Log.w
- 巧用键盘的快捷键,提高工作效率
- Oracle常用脚本——监控数据库中的活跃用户及其运行
- android中使用startActivityForResult回传数据
- Android开发者应该深入学习的10个开源应用项目
- cocos2dx使用project-creator生成工程文件
- arguments对象解析
- Android模拟器无法上网问题