HUD 3706 单调队列简单题

Problem Description

Give you three integers n, A and B.

Then we define S_i = Aⁱ mod B and T_i = Min{ S_k | i-A <= k <= i, k >= 1}

Your task is to calculate the product of T_i (1 <= i <= n) mod B.

不描述题意了，三行英文挺明了的，今天刚学单调队列，自己模拟一下过了这题

单调队列

1，永远保持队首元素为最小值。

2，插入时，先将队尾大于插入值的元素删去

3，保证队首元素一直是合法，所需元素。

代码

#include<stdio.h>

#include<math.h>

int tou, wei, count;

struct cc

{

__int64 value;

int num;

}mark[10000005];

void add ( int x ,int k)

{

if(k == 1)  return;

if( tou == wei )  {mark[tou].value = x; mark[wei].num = k; return;}

if(mark[tou].value > x) {mark[tou].value = x; mark[tou].num = k; wei = tou+1; return;}

for(int i = wei - 1; i >= tou; i--)

{

if(mark[i].value <= x)

{

mark[i+1].value = x;

mark[i+1].num = k;

wei = i + 2;

break;

}

}

for(int i = tou; i < wei; i++)

{

if(mark[i].num < count)

{

tou++;

}

else break;

}

}

int main()

{

int n, a, b;

__int64 k;

__int64 sum, ji;

while(scanf("%d%d%d", &n, &a, &b) != EOF)

{

count = -1, ji = 1;

mark[0].value = a % b;

mark[0].num = 1;

tou = 0, wei = 1, k = 1;

for(int i = 1; i <= n; i++)

{

k = k * a % b;

count = i - a;

add(k, i);

ji *= mark[tou].value;

if( ji >= b ) ji %= b;

}

printf("%I64d\n", ji);

}

}