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;
}