http://acm.hdu.edu.cn/showproblem.php?pid=5475

An easy problem

Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 755 Accepted Submission(s):
431

Problem Description

One day, a useless calculator was being built by Kuros. Let's assume that number X is showed on the screen of calculator. At first, X = 1. This calculator only supports two types of operation.
1. multiply X with a
number.
2. divide X with a number which was multiplied before.
After each
operation, please output the number X modulo M.

Input

The first line is an integer T(1≤T≤10^5), indicating the number of test cases.

For each test case, the first line
are two integers Q and M. Q is the number of operations and M is described
above. (1≤Q≤10^5,1≤M≤10^9)

The next Q lines, each line starts with an integer x indicating the type of
operation.

if x is 1, an integer y is given, indicating the number to multiply. (0<y≤10^9)
if x is 2, an integer n is given. The calculator will divide the number
which is multiplied in the nth operation. (the nth operation must be a type 1
operation.)

It's guaranteed that in type 2 operation, there won't be two
same n.

Output

For each test case, the first line, please output "Case #x:" and x is the id of the test cases starting from 1.
Then Q lines follow,
each line please output an answer showed by the calculator.

Sample Input

10 1000000000

1 2

2 1

1 2

1 10

2 3

2 4

1 6

1 7

1 12

2 7

Sample Output

Case #1:

504

题目大意：t组测试数据，q个操作，每个操作（a, y）的结果X对m取余，X最初值为1，如果a==1，将X乘以y

如果a==2，将X除以第y次操作的数（即X要除的数曾出现在乘操作里）

线段树：线段树区间维护乘积，将q次操作当做q个节点，每个区间的初始值是1，乘操作（a，y）：将该次操作即该节点的数值改成其要乘的值y，

除操作（a，y）：将第y次操作即y节点的数值改成1（即可让其免去乘操作），输出从1到q区间的乘积即可

#include<stdio.h>

#include<string.h>

#include<math.h>

#include<stdlib.h>

#include<algorithm>

#define Lson root<<1, L, tree[root].Mid()

#define Rson root<<1|1, tree[root].Mid() + 1, R

const int N = ;

typedef long long ll;

struct Tree

{

    ll L, R;

    ll sum;

    int Mid()

    {

        return (L + R) / ;

    }

} tree[N * ];

ll a[N], m;

void Push(int root)

{

    tree[root].sum = (tree[root<<].sum * tree[root<<|].sum) % m;

}//维护区间乘积

void Build(int root, ll L, ll R)

{

    tree[root].L = L, tree[root].R = R;

    if(L == R)

    {

        tree[root].sum = ;

        return ;

    }

    Build(Lson);

    Build(Rson);

    Push(root);

}//建树

void Update(int root, ll op, ll e)

{

    if(tree[root].L == op && tree[root].R == op)

    {

        tree[root].sum = e % m;

        return ;

    }

    if(op <= tree[root].Mid())

        Update(root<<, op, e);

    else

        Update(root<<|, op, e);

    Push(root);

}//区间单点更新

int main()

{

    int t, q, op, x = ;

    scanf("%d", &t);

    while(t--)

    {

        x++;

        scanf("%d%lld", &q, &m);

        printf("Case #%d:\n", x);

        Build(, , q);

        for(int i =  ; i <= q ; i++)

        {

            scanf("%d%lld", &op, &a[i]);

            if(op == )

                Update(, i, a[i]);

            else

                Update(, a[i], );

            printf("%lld\n", tree[].sum);

        }

    }

    return ;

}

通过用线段树的思路可以用暴力的方法来解题

#include<stdio.h>

#include<string.h>

#include<math.h>

#include<stdlib.h>

#include<algorithm>

#define Lson root<<1, L, tree[root].Mid()

#define Rson root<<1|1, tree[root].Mid() + 1, R

const int N = ;

typedef long long ll;

ll a[N], m;

int main()

{

    int t, op, q, x = ;

    scanf("%d", &t);

    while(t--)

    {

        x++;

        ll ans = ;

        scanf("%d%lld", &q, &m);

        printf("Case #%d:\n", x);

        for(int i =  ; i <= q ; i++)

        {

            scanf("%d%lld", &op, &a[i]);

            if(op == )

                ans = ans * a[i] % m;

            else

            {

                a[a[i]] = ;

                a[i] = ;

                ans = ;

                for(int j =  ; j < i ; j++)

                    ans = ans * a[j] % m;

            }

            printf("%lld\n", ans);

        }

    }

    return ;

}

巴特西

hdu 5475 An easy problem（暴力 || 线段树区间单点更新）

An easy problem

最新文章

热门文章