HDU-5475 An easy problem（模拟||(倒着计算+线段树)）

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最新推荐文章于 2020-07-21 21:21:14 发布

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分类专栏：其他线段树 HDU 文章标签： HDU 模拟线段树

本文链接：https://blog.youkuaiyun.com/idealism_xxm/article/details/48756091

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本文介绍了一道关于计算器操作的编程题，包括题目描述、输入输出格式及样例。通过两种解题思路对比，详细解释了如何高效地处理乘除操作，并给出具体代码实现。

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An easy problem

Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

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), 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≤105,1≤M≤109)
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≤109)
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

1
10 1000000000
1 2
2 1
1 2
1 10
2 3
2 4
1 6
1 7
1 12
2 7

Sample Output

题目大意：初始时 x=1，每次有2种操作：

操作 1：给x乘以一个数

操作 2：给x除以第n次操作出现的数（n只出现1次）

对每次操作输出 x%mod;

①刚开始就想到直接计算，但认为会超时就直接放弃了，没想到直接计算（乘法直接乘即可；除法时先标记除法，然后重新计算即可）就能过，大概3400ms；

②后来听过大神讲解，了解到：有些数一定会乘，计算为tmp[i]，然后从后计算除法除以的数n[i]，则ans[i]=(tmp[i]*n[i])%mod;，大概2300ms；

#include <cstdio>
#define LL long long

using namespace std;

int n[100005],ope[100005],tmp[100005],ans[100005],x,mod;
bool flg[100005],mul[100005];//flg表示当前操作是否为操作1；mul表示当前数是否一定会被乘

int main() {
    int T,kase=0,Q,t,i,j;
    scanf("%d",&T);
    while(kase<T) {
        printf("Case #%d:\n",++kase);
        scanf("%d%d",&Q,&mod);
        tmp[0]=1;
        for(i=1;i<=Q;++i) {
            scanf("%d%d",&t,ope+i);
            n[i]=1;
            if(t==1)
                mul[i]=flg[i]=true;
            else
                flg[i]=mul[i]=mul[ope[i]]=false;
        }
        for(i=1;i<=Q;++i) {
            if(mul[i])
                tmp[i]=((LL)tmp[i-1]*ope[i])%mod;
            else
                tmp[i]=tmp[i-1];
        }
        for(i=Q;i;--i) {
            if(!flg[i]) {
                x=ope[ope[i]];
                for(j=ope[i];j<i;++j)//ope[i]~i-1的n[j]均乘以x
                    n[j]=((LL)n[j]*x)%mod;
            }
            ans[i]=((LL)n[i]*tmp[i])%mod;
        }
        for(i=1;i<=Q;++i)
            printf("%d\n",ans[i]);
    }
    return 0;
}

③大神自己用线段树做的，大概1500ms。