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Orz iwtwiioi
学习了一下用记忆化搜索来捉题的新姿势……但没学会TAT,再挖个坑(妈蛋难道对我来说数位DP就是个神坑吗……sigh)
1 //BZOJ 1833 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cstring> 5 #include<iostream> 6 #include<algorithm> 7 #define rep(i,n) for(int i=0;i<n;++i) 8 #define F(i,j,n) for(int i=j;i<=n;++i) 9 #define D(i,j,n) for(int i=j;i>=n;--i) 10 using namespace std; 11 typedef long long LL; 12 LL getint(){ 13 LL v=0,sign=1; char ch=getchar(); 14 while(ch<‘0‘||ch>‘9‘) {if (ch==‘-‘) sign=-1; ch=getchar();} 15 while(ch>=‘0‘&&ch<=‘9‘) {v=v*10+ch-‘0‘; ch=getchar();} 16 return v*=sign; 17 } 18 /*******************tamplate********************/ 19 20 21 LL f[100],c[100],a[100],p[100]; 22 LL dfs(int x,int dig,int front,int line){ 23 if (!x) return 0; 24 if (!front && !line && f[x]!=-1) return f[x]; 25 LL last=(line ? a[x] : 9), tot=0; 26 F(i,0,last){ 27 if (front && i==0) tot+=dfs(x-1,dig,1,line&&i==last); 28 else if (i==dig){ 29 if (i==last && line) tot+=c[x-1]+1+dfs(x-1,dig,0,line && i==last); 30 else tot+=p[x-1]+dfs(x-1,dig,0,line && i==last); 31 } 32 else tot+=dfs(x-1,dig,0,line&& i==last); 33 } 34 if (!front && !line) f[x]=tot; 35 return tot; 36 } 37 LL getans(LL x,LL dig){ 38 memset(f,-1,sizeof f); 39 LL t=x; int len=0; 40 while(t) a[++len]=t%10,t/=10,c[len]=c[len-1]+a[len]*p[len-1]; 41 return dfs(len,dig,1,1); 42 } 43 int main(){ 44 // freopen("input.txt","r",stdin); 45 LL a=getint(),b=getint(); 46 p[0]=1; F(i,1,15) p[i]=p[i-1]*10; 47 rep(i,9) printf("%lld ",getans(b,i)-getans(a-1,i)); 48 printf("%lld\n",getans(b,9)-getans(a-1,9)); 49 return 0; 50 }
【BZOJ】【1833】【ZJOI2010】count 数字计数
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原文地址:http://www.cnblogs.com/Tunix/p/4309858.html