Integer Approximation Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5081   Accepted: 1652 Description The FORTH programming language does not support floating-point arithmetic at all. Its author, Chuck Moore, maintains that floating-po…

1.链接地址: http://poj.org/problem?id=1503 2.题目: Integer Inquiry Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 28115   Accepted: 10925 Description One of the first users of BIT's new supercomputer was Chip Diller. He extended his explorati…

Integer Intervals Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12192   Accepted: 5145 Description An integer interval [a,b], a < b, is a set of all consecutive integers beginning with a and ending with b. Write a program that: finds t…

题目链接:http://poj.org/problem?id=1503 import java.io.*; import java.math.BigInteger; import java.util.Scanner; public class Main { public static void main(String []args) throws IOException{ Scanner scanner=new Scanner(System.in); BigInteger a=BigIntege…

题目:http://poj.org/problem?id=1716 #include <stdio.h> #include <string.h> #include <vector> #include <queue> const int INF = 0x3f3f3f3f; struct Edge { int v, w; Edge(){}; Edge(int v, int w) { this->v = v; this->w = w; } }; std…

题目链接:http://poj.org/problem?id=1503 思路分析: 基本的高精度问题,使用字符数组存储然后处理即可. 代码如下: #include <iostream> #include <string> using namespace std; + ; char input[M]; int A[M], sum[M]; void Reverse(int A[], const char str[]) { int len = strlen(str); ; ; i >…

一.Description One of the first users of BIT's new supercomputer was Chip Diller. He extended his exploration of powers of 3 to go from 0 to 333 and he explored taking various sums of those numbers. ``This supercomputer is great,'' remarked Chip. ``I…

POJ 1201 http://poj.org/problem?id=1201 题目大意: 有一个序列,题目用n个整数组合 [ai,bi,ci]来描述它,[ai,bi,ci]表示在该序列中处于[ai,bi]这个区间的整数至少有ci个.如果存在这样的序列,请求出满足题目要求的最短的序列长度是多少. 思路: 设s[i]为从1~i的整数个数. 这样对于区间[ a , b]显然有 S[b+1] - S[a] >=c[i] (为什么是b+1?因为闭区间b也要选上呀) 然后还有 0<= S[B+1]-S[…

1716 -- Integer Intervals 跟之前个人赛的一道二分加差分约束差不多,也是求满足条件的最小值. 题意是,给出若干区间,需要找出最少的元素个数,使得每个区间至少包含两个这里的元素. 做法就是建立(b)->(a)=-2,(i)->(i+1)=1,(i+1)->(i)=0的边,然后跑一次spfa即可. 做完的时候,因为队列开太小,所以re了一次. 代码如下: #include <iostream> #include <algorithm> #inc…

题目 我要开始练习一些java的简单编程了^v^ import java.io.*; import java.util.*; import java.math.*; public class Main { /** * @param args */ public static void main(String[] args) throws Exception { // 定义并打开输入文件 Scanner cin = new Scanner(System.in); BigInteger a, sum…