Romantic Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 2669 Description The Sky is Sprite. The Birds is Fly in the Sky. The Wind is Wonderful. Blew Throw the Trees Trees are Shaking, Leaves ar…

Romantic Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2835    Accepted Submission(s): 1117 Problem Description The Sky is Sprite.The Birds is Fly in the Sky.The Wind is Wonderful.Blew Throw t…

Romantic Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2385    Accepted Submission(s): 944 Problem Description The Sky is Sprite.The Birds is Fly in the Sky.The Wind is Wonderful.Blew Throw th…

Romantic Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6643    Accepted Submission(s): 2772 Problem Description The Sky is Sprite.The Birds is Fly in the Sky.The Wind is Wonderful.Blew Throw t…

Romantic Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8536    Accepted Submission(s): 3638 Problem Description The Sky is Sprite.The Birds is Fly in the Sky.The Wind is Wonderful.Blew Throw t…

题目链接:pid=2669">http://acm.hdu.edu.cn/showproblem.php?pid=2669 Problem Description The Sky is Sprite. The Birds is Fly in the Sky. The Wind is Wonderful. Blew Throw the Trees Trees are Shaking, Leaves are Falling. Lovers Walk passing, and so are Yo…

裸的扩展欧几里德,求最小的X,X=((X0%b)+b)%b,每个X都对应一个Y,代入原式求解可得 #include<stdio.h> #include<string.h> typedef long long ll; ll ex_gcd(ll a,ll b,ll &x,ll &y){ if(!b){ x=,y=; return a; } int ans=ex_gcd(b,a%b,y,x); y-=a/b*x; return ans; } void cal(ll a,l…

题目 //第一眼看题目觉得好熟悉,但是还是没想起来//洪湖来写不出来去看了解题报告,发现是裸的 扩展欧几里得 - - /* //扩展欧几里得算法(求 ax+by=gcd )//返回d=gcd(a,b);和对应于等式ax+by=d中的x,y#define LL long longLL extend_gcd(LL a,LL b,LL &x,LL &y){ if(a==0&&b==0) return -1;//无最大公约数 if(b==0){x=1;y=0;return a;}…

Now tell you two nonnegative integer a and b. Find the nonnegative integer X and integer Y to satisfy X*a + Y*b = 1. If no such answer print "sorry" instead. InputThe input contains multiple test cases.Each case two nonnegative integer a,b (0<…

链接:传送门 题意:求解方程 X * a + Y * b = 1 的一组最小非负 X 的解,如果无解输出 "sorry" 思路:裸 exgcd /************************************************************************* > File Name: hdu2669.cpp > Author: WArobot > Blog: http://www.cnblogs.com/WArobot/ > C…