列表

详情


NC51409. subsequence 1

描述

You are given two strings s and t composed by digits (characters '0' '9'). The length of s is n and the length of t is m. The first character of both s and t aren't '0'.

Please calculate the number of valid subsequences of s that are larger than t if viewed as positive integers. A subsequence is valid if and only if its first character is not '0'.
Two subsequences are different if they are composed of different locations in the original string. For example, string "1223" has 2 different subsequences "23".

Because the answer may be huge, please output the answer modulo 998244353.

输入描述

The first line contains one integer T, indicating that there are T tests.

Each test consists of 3 lines.

The first line of each test contains two integers n and m, denoting the length of strings s and t.

The second line of each test contains the string s.

The third line of each test contains the string t.

* .

* sum of n in all tests .

* the first character of both s and t aren't '0'.

输出描述

For each test, output one integer in a line representing the answer modulo 998244353.

示例1

输入:

3
4 2
1234
13
4 2
1034
13
4 1
1111
2

输出:

9
6
11

说明:

For the last test, there are 6 subsequences "11", 4 subsequcnes "111" and 1 subsequence "1111" that are valid, so the answer is 11.

原站题解

上次编辑到这里,代码来自缓存 点击恢复默认模板

C++(clang++ 11.0.1) 解法, 执行用时: 110ms, 内存消耗: 63396K, 提交时间: 2022-12-06 12:05:34

#include<stdio.h>
#include<string.h>
const int mo=998244353;
char a[3010],b[3010];
int c[3010][3010],dp[3010][3010];
int main(){
	int t,n,m,i,j;
	scanf("%d",&t);
	for(i=0;i<=3000;i++)
	{
		c[i][0]=1;
		for(j=1;j<=i;j++)
		c[i][j]=(c[i-1][j-1]+c[i-1][j])%mo;
	}
	while(t--)
	{
		scanf("%d%d",&n,&m);
		scanf("%s%s",a+1,b+1);
		long long ans=0;
		for(i=1;i<=n;i++)
		if(a[i]!='0')
		for(j=m;j<=n-i;j++)
		(ans+=(c[n-i][j]))%=mo;
		for(i=0;i<=n;i++)
		for(j=0;j<=m;j++)
		dp[i][j]=0;
		dp[0][0]=1;
		for(i=1;i<=n;i++)
		{
			dp[i][0]=1;
			for(j=1;j<=m;j++)
			{
				dp[i][j]=dp[i-1][j];
				if(a[i]==b[j])
				(dp[i][j]+=dp[i-1][j-1])%=mo;
				else if(a[i]>b[j])
				ans=(ans+1ll*dp[i-1][j-1]*c[n-i][m-j])%mo;
			}
		}
		printf("%lld\n",ans);
	}
	return 0;
}

C++14(g++5.4) 解法, 执行用时: 250ms, 内存消耗: 117576K, 提交时间: 2019-08-02 11:11:16

#include <bits/stdc++.h>
#define rep(i, a, b) for (register int i = a; i <= b; i++)
using namespace std;
typedef long long ll;
const int mod = 998244353;

ll a[3010][3010];
int n,m;
char s[3010],t[3010];
ll ans=0;
ll dp[3010][3010];
int main() {
	int T;
	cin>>T;
	a[0][0]=1;
	rep(i,1,3005)rep(j,0,i)a[i][j]=(a[i-1][j-1]+a[i-1][j])%mod;
	while(T--){
		cin>>n>>m;
		cin>>s>>t;
		ans=0;
		rep(i,0,n-m-1)if(s[i]!='0')rep(j,m,n-i-1)ans=(ans+a[n-i-1][j])%mod;
		rep(i,0,n)rep(j,0,m)dp[i][j]=0;
		for(int j=m-1;j>=0;j--){
			for(int i=n-m+j;i>=0;i--){
				if(s[i]==t[j])
					dp[i][j]=dp[i+1][j+1];
				else if(s[i]>t[j])
					dp[i][j]=a[n-i-1][m-j-1];
				dp[i][j]=(dp[i][j]+dp[i+1][j])%mod;
			}
		}
		cout<<(ans+dp[0][0])%mod<<endl;
	}
	return 0;    
}

上一题