sdut 1446 超级玛丽

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=1446超级玛丽

题意：

中文...

 思路：

比赛时，数据弱了，让我一个O（10^12）的程序都过了，后来就没多想，加上数据后。改为O(n)的才过。枚举能跳过的云彩数，然后对跳跃的长度D取余求商，计算余数+ M与L的差值（即云彩之间的距离）在检查看看能否跳过，如果不能，就截止在这里不会继续往下跳了。

注意数据类型long long

//#pragma comment(linker,"/STACK:327680000,327680000")  
#include <iostream>  
#include <cstdio>  
#include <cmath>  
#include <vector>  
#include <cstring>  
#include <algorithm>  
#include <string>  
#include <set>  
#include <functional>  
#include <numeric>  
#include <sstream>  
#include <stack>  
#include <map>  
#include <queue>  
  
#define CL(arr, val)    memset(arr, val, sizeof(arr))  
  
#define ll long long  
#define inf 0x7f7f7f7f  
#define lc l,m,rt<<1  
#define rc m + 1,r,rt<<1|1  
#define pi acos(-1.0)  
#define ll long long  
#define L(x)    (x) << 1  
#define R(x)    (x) << 1 | 1  
#define MID(l, r)   (l + r) >> 1  
#define Min(x, y)   (x) < (y) ? (x) : (y)  
#define Max(x, y)   (x) < (y) ? (y) : (x)  
#define E(x)        (1 << (x))  
#define iabs(x)     (x) < 0 ? -(x) : (x)int  
#define OUT(x)  printf("%I64d\n", x)  
#define lowbit(x)   (x)&(-x)  
#define Read()  freopen("din.txt", "r", stdin)  
#define Write() freopen("dout.txt", "w", stdout);  
  
  
  
using namespace std;  
  
ll N[2],D[2],M[2],L[2];  
  
int main()  
{  
    int i;  
    ll pos1,pos2;  
    ll ct1,ct2 ;  
    int T;  
    scanf("%d",&T);  
    while (T--)  
    {  
        ct1 = ct2 = 0;  
        cin>>N[0]>>D[0]>>M[0]>>L[0];  
        cin>>N[1]>>D[1]>>M[1]>>L[1];  
  
        ll sh = 0, yu = 0;  
        for (i = 0; i < N[0]; ++i)  //枚举跳过云彩的数量
        {  
            pos1 = i*M[0] + L[0];  //计算长度
            sh = pos1/D[0];  //取余求商
            yu = pos1%D[0];  
            pos1 = sh*D[0] + D[0];  //跳了cnt1后的距离
            ll s = yu + M[0] - L[0];  //这是关键来决定是否能够继续往下跳
            if (s > D[0])  
            {  
                ct1 = pos1/D[0];  
                break;  
            }  
        }  
        if (i >= N[0])  
        {  
            pos1 = sh*D[0] + D[0];  
            ct1 = pos1/D[0];  
        }  
  
  
        for (i = 0; i < N[1]; ++i)  
        {  
            pos2 = i*M[1] + L[1];  
            sh  = pos2/D[1];  
            yu = pos2%D[1];  
            pos2 = sh*D[1] + D[1];  
  
            ll s = yu + M[1] - L[1];  
            if (s > D[1])  
            {  
                ct2 = pos2/D[1];  
                break;  
            }  
        }  
        if (i >= N[1])  
        {  
            pos2 = sh*D[1] + D[1];  
            ct2 = pos2/D[1];  
  
        }  
  
        if (ct1 == ct2)  
        {  
            printf("Az is Winner at %lld\n",pos2);  
        }  
        else  
        {  
            ll ans = 0;  
            if (ct1 > ct2) ans = pos1;  
            else ans = pos2;  
            printf("Lz is Winner at %lld\n",ans);  
        }  
    }  
    return 0;  
}