@xunuo
2017-01-16T19:29:53.000000Z
字数 1233
阅读 1092
Time limit 4000 ms Memory limit 65536 kB
单调队列
来源:
vjudge:B-Second My Problem First
HDU:HDU 3706 Second My Problem First
Give you three integers n, A and B.
Then we define Si = A^i mod B and Ti = Min{ Sk | i-A <= k <= i, k >= 1}
Your task is to calculate the product of Ti (1 <= i <= n) mod B.
Each line will contain three integers n(1 <= n <= 107),A and B(1 <= A, B <= 231-1).
Process to end of file.
For each case, output the answer in a single line.
1 2 3
2 3 4
3 4 5
4 5 6
5 6 7
2
3
4
5
6
题意:
其实最开始根本就不懂这个题什么意思,然而找题解的时候都没有说题意!!!都说直接看题就好,意思很好懂,,,我的天哪!!!你自己还是太年轻!!现在来说一下题意:
首先输入会有三个数:n,a,b;
题目的意思是让你最终输出:Ti%b,而Ti是由这句话得到:
Ti=Min{Sk| i-A<=k<=i,k>=1}
表示的是Sk的最小值,而k的范围是i-A<=k<=i,k给出来的是>=1的,不是需要你去判断它是否>=1!!!是你自己zz了
而Si=A^i%b;
i是从1到n!!!
解题思路:
看懂题之后,就只需要用到单调递减队列来维护一个Si的最大值就好!但是要注意k的范围;
完整代码:
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define ll long long
using namespace std;
ll q[100010];
ll p[100010];
///用题中的10^7要MLE,改小了数组居然过了,我并不知道why!!!
int main()
{
ll n,a,b;
ll s,ans;
ll head,tail;
while(scanf("%lld%lld%lld",&n,&a,&b)!=EOF)
{
memset(q,0,sizeof(q));
memset(p,0,sizeof(p));
s=1,ans=1;
head=0,tail=0;
for(int i=1;i<=n;i++)
{
s=(s*a)%b;
while(head<tail&&q[tail-1]>s)///维护题中Ti的最小值;
tail--;
q[tail]=s;
p[tail]=i;
tail++;
while(head<tail&&i-a>p[head])///要满足;i-A<=k<=i
head++;
ans=(ans*q[head])%b;
}
printf("%lld\n",ans);
}
return 0;
}