https://www.lydsy.com/JudgeOnline/problem.php?id=1833

題解：數(shù)位DP

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; ll t,n,m,k,q; int ans,cnt,flag,temp; ll f[100], c[100], a[100], p[100];ll dfs(int x, int dig, int front, int limit) {if(!x) return 0;if(!front && !limit && f[x]!=-1) return f[x];ll last=(limit?a[x]:9), tot=0;for(int i=0;i<=last;i++) {if(front && i==0)tot+=dfs(x-1, dig, 1, limit&&i==last);else if(i==dig) {if(i==last && limit)tot+=c[x-1]+1+dfs(x-1, dig, 0, limit&&i==last); //正好在這個數(shù)上elsetot+=p[x-1]+dfs(x-1, dig, 0, limit&&i==last);}else tot+=dfs(x-1, dig, 0, limit&&i==last);}if(!front && !limit) f[x]=tot;return tot; } ll getans(ll x, int dig) {memset(f,-1,sizeof(f));ll t=x; int len=0;while(t) a[++len]=t%10, t/=10, c[len]=c[len-1]+a[len]*p[len-1];return dfs(len, dig, 1, 1); } int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//scanf("%d",&n);//scanf("%d",&t);while(~scanf("%lld%lld",&n,&m)){p[0]=1; for(int i=1;i<15;i++) p[i]=p[i-1]*10;for(int i=0;i<9;i++)printf("%lld ", getans(m, i)-getans(n-1, i));printf("%lld\n", getans(m, 9)-getans(n-1, 9));}//cout << "Hello world!" << endl;return 0; }

?

總結

以上是生活随笔為你收集整理的count 数字计数的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內容還不錯，歡迎將生活随笔推薦給好友。