skyfire1123的博客

http://www.patest.cn/contests/pat-a-practise/1010 思路：都转化为10进制，然后比较大小即可。 BUG之处在于：题目中测试用的进制可能非常的大，表现为：如果把下列程序的max缩小，错误测试点会增加，而如果增加到很大，个别测试点会超时，并且long long也未必能存下这么大的数字，待学过算法以后再来解决。#include <iostream>#i

http://www.patest.cn/contests/pat-a-practise/1011#include <stdio.h>const char s[]="WTL";int main(){ float bet[3][3]; for(int i=0;i<3;++i){ for(int j=0;j<3;++j){ scanf("%f"

http://www.patest.cn/contests/pat-a-practise/1008#include <iostream>using namespace std;int elevator_go(int,int);int main(){ int N,nex,pre=0,sum=0; cin>>N; while(N--){ cin>>nex;

http://www.patest.cn/contests/pat-a-practise/1005#include <iostream>#include <string>#include <cmath>int main(){ using namespace std; const string Num[]={"zero","one","two","three","four","f

//http://www.patest.cn/contests/pat-a-practise/1007#include #define maxr 55000#define maxn 10005using namespace std;int seq[maxn];int sum_seq(int i,int j);int main(){ int K,cnt=0; int sum[m

#include <iostream>#include <string>#define maxr 100using namespace std;string record[maxr][3];int main(){ unsigned int M; cin>>M; for(int i=0;i<M;++i){ cin>>record[i][0]>>recor

#include <stdio.h>#define maxn 1005int main(){ int K; float NK[maxn]={0}; while(scanf("%d",&K)!=EOF){ while(K--){ int i; float temp=0.0; scanf(

#include <stdio.h>int main(){ int a,b; scanf("%d%d",&a,&b); int sum =a+b; if(sum<0){ printf("-"); sum=-sum; } if(sum>=1000000){ printf("%d,%03d,%03d\n",

http://www.patest.cn/contests/pat-a-practise/1023注意点：题目中的数不超过20位，而64位long long整型只有19位，所以这里采用字符数组的形式保存整数的每一位，难点在于字符数组乘以2怎么实现。#include #define maxn 25using namespace std;int main(){ int a[10]=

http://www.patest.cn/contests/pat-a-practise/1019#include using namespace std;void print(int *a,int digit){ for(int i=digit-1;i>0;--i){ cout<<a[i]<<' '; } cout<<a[0];}int main(){ int N,

http://www.patest.cn/contests/pat-a-practise/1015 思路：用数组存储转化进制以后的各位，既可以避免溢出，也方便”反转“。#include <iostream>#include <cmath>using namespace std;bool is_prime(int num){ if(num<2){ return false

#include <iostream>#include <string>using namespace std;int N,M;struct stu_grade{ //包含学生的学号和各科成绩的结构体 string ID; unsigned int C,M,E,A;};void bestrank_print(int,s

http://www.patest.cn/contests/pat-a-practise/1009 思路：用数组的下标对应指数，数组储存的值对应系数。记录每个多项式的最高次以减少运算量。#include <stdio.h>#define maxn 1005#define LOCALusing namespace std;int main(){#ifdef LOCAL freope