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http://poj.org/problem?id=1068这个题的话就是先把给出来的一串数字转化成括号,再把括号转化成要求的,最后输出就行了#include#include#include#include#include#includeusing namespace std ;int main(){ stackQ; queuep,q; int n; cin>>n; for(int i = 1 ; i >m; int a[66]; for(int j = 1 ; j >a[j] ; if(j == 1) ... 阅读全文
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Alice and BobTime Limit: 1000ms Memory limit: 65536K题目描述 Alice and Bob like playing games very much.Today, they introduce a new game. There is apolynomial like this:(a0*x^(2^0)+1) * (a1* x^(2^1)+1)*.......*(an-1* x^(2^(n-1))+1). Then Alice ask Bob Q questions. In theexpansion of the Polynomial, Gi.. 阅读全文
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做模拟题做的我直接睡着了,题并不难,就是一个细心的问题,有一些细节问题注意了就差不多了,代码写的精美的一般找错误也好找一些,应该学着些好看的代码 1 #include 2 #include 3 #include 4 using namespace std ; 5 int len,high; 6 int a,b,j ; 7 struct node 8 { 9 int x; 10 int y ; 11 int dire; 12 } s[10001]; 13 int judge(int k) 14 { 15 int i; 16 if(s[k]... 阅读全文