CodeForces - 1567C Carrying Conundrum(思维/状压)

Posted 2022-02-23 Frozen_Guardian

题目链接：点击查看

题目大意：规定加法中使用隔项进位，问给定的 $n$ 有多少种方案可以通过 “隔项进位加法” 得到

题目分析：隔项进位意味着奇偶位置的数字互不影响，所以将奇偶位置的两个数字拿出来，设为 $a$ 和 $b$。不难看出组合出某个数字 $x$ 的方案为 $x+1$，即 $0+(x),1+(x-1),\cdots ,x+(0)$。所以总方案数为 $(a+1)\ast (b+1)$，最终答案需要减去二，因为组合出来的数必须非零。

因为时间限制比较宽裕，所以本题可以直接状压枚举哪些位置是进位的，然后统计就好了

代码：

// Problem: C. Carrying Conundrum
// Contest: Codeforces - Codeforces Round #742 (Div. 2)
// URL: https://codeforces.com/contest/1567/problem/C
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast","inline","-ffast-math")
// #pragma GCC target("avx,sse2,sse3,sse4,mmx")
#include<iostream>
#include<cstdio>
#include<string>
#include<ctime>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<climits>
#include<queue>
#include<map>
#include<set>
#include<sstream>
#include<cassert>
#include<bitset>
#include<list>
#include<unordered_map>
#define lowbit(x) (x&-x)
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
template<typename T>
inline void read(T &x)

	T f=1;x=0;
	char ch=getchar();
	while(0==isdigit(ch))if(ch=='-')f=-1;ch=getchar();
	while(0!=isdigit(ch)) x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
	x*=f;

template<typename T>
inline void write(T x)

	if(x<0)x=~(x-1);putchar('-');
    if(x>9)write(x/10);
    putchar(x%10+'0');

const int inf=0x3f3f3f3f;
const int N=1e6+100;
int main()

#ifndef ONLINE_JUDGE
//	freopen("data.in.txt","r",stdin);
//	freopen("data.out.txt","w",stdout);
#endif
//	ios::sync_with_stdio(false);
	int w;
	cin>>w;
	while(w--) 
		string s;
		cin>>s;
		int ans1=0,ans2=0;
		for(int i=0;i<(int)s.size();i++) 
			if(i&1) 
				ans1=ans1*10+s[i]-'0';
			 else 
				ans2=ans2*10+s[i]-'0';
			
		
		cout<<1LL*(ans1+1)*(ans2+1)-2<<endl;
	
	return 0;

// Problem: C. Carrying Conundrum
// Contest: Codeforces - Codeforces Round #742 (Div. 2)
// URL: https://codeforces.com/contest/1567/problem/C
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast","inline","-ffast-math")
// #pragma GCC target("avx,sse2,sse3,sse4,mmx")
#include<iostream>
#include<cstdio>
#include<string>
#include<ctime>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<climits>
#include<queue>
#include<map>
#include<set>
#include<sstream>
#include<cassert>
#include<bitset>
#include<list>
#include<unordered_map>
#define lowbit(x) (x&-x)
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
template<typename T>
inline void read(T &x)

	T f=1;x=0;
	char ch=getchar();
	while(0==isdigit(ch))if(ch=='-')f=-1;ch=getchar();
	while(0!=isdigit(ch)) x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
	x*=f;

template<typename T>
inline void write(T x)

	if(x<0)x=~(x-1);putchar('-');
    if(x>9)write(x/10);
    putchar(x%10+'0');

const int inf=0x3f3f3f3f;
const int N=1e6+100;
int sum[N];
void init() 
	for(int i=0;i<10;i++) 
		for(int j=0;j<10;j++) 
			sum[i+j]++;
		
	

int main()

#ifndef ONLINE_JUDGE
//	freopen("data.in.txt","r",stdin);
//	freopen("data.out.txt","w",stdout);
#endif
//	ios::sync_with_stdio(false);
	init();
	int w;
	cin>>w;
	while(w--) 
		int n;
		read(n);
		LL ans=-2;
		for(int i=0;i<1<<8;i++) 
			int res=1,nn=n;
			for(int j=0;j<10;j++,nn/=10) 
				int cur=nn%10;
				if(i>>j&1) 
					cur+=10;
				
				if(j>=2&&((i>>(j-2))&1)) 
					cur--;
				
				if(cur<0) 
					res=0;
					break;
				
				res*=sum[cur];
			
			ans+=res;
		
		cout<<ans<<endl;
	
	return 0;