数位dp n内1的个数递推找规律

1061:数字统计

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总时间限制: 

1000ms 

内存限制: 

10000kB

描述

给出一个整数n(1<=n<=20000000),要求输出从1到n间所有数字中“1”的出现次数．例如：数字11，1到11间数字“1”的出现次数为4。（1，10，11，11出现4次，因为11有两个1，所以出现4次）

输入

一个整数n，(1<=n<=20000000)

输出

输出一行，输出“1”的出现次数。

样例输入

11

样例输出

4

http://wyu.openjudge.cn/practice/1061

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<vector>#include<queue>#include<map>#include<stack>#define rt return#define bk break#define ct continue#define sf scanf#define pf printf#define ms memset#define si(n) sf("%d",&n)#define pi(n) pf("%d\n",n)#define REP0(i,n) for(int i=0;i<(n);i++)#define REP1(i,n) for(int i=1;i<=(n);i++)#define REP(i,s,n) for(int i=s;i<=(n);i++)#define db double#define op operator#define pb push_back#define LL long long#define INF 0x3fffffff#define eps 1e-8#define PI acos(-1)#define maxn 1010using namespace std;int f[100];void init(){    f[0]=0;    for(int i=1;i<=9;i++)f[i]=1;    f[10]=2;    f[11]=4;    for(int i=12;i<=19;i++)f[i]=f[i-1]+1;    for(int i=20;i<=99;i++){        if(i%10==1)f[i]=f[i-1]+1;        else f[i]=f[i-1];    }}int dp(int n){    if(n<=99)rt f[n];    int tmp=1;    while(tmp<=n)tmp*=10;    tmp/=10;    int res;    if(n/tmp==1){        res=dp(tmp-1)+dp(n%tmp)+10*(n/10-tmp/10)+(n%10)+1;    }else {        res=(n/tmp)*dp(tmp-1)+dp(n%tmp)+tmp;    }    rt res;}int main(){    #ifdef ACBang    freopen("in.txt","r",stdin);    freopen("dp.txt","w",stdout);    #endif    init();    int n;    while(~sf("%d",&n)){        int ans;        if(n<=99)ans=f[n];        else ans=dp(n);       // pf("%d %d\n",n,ans);       pf("%d\n",ans);    }    rt 0;}

//修剪版！！！！！！！！！

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<vector>#include<queue>#include<map>#include<stack>#define rt return#define bk break#define ct continue#define sf scanf#define pf printf#define ms memset#define si(n) sf("%d",&n)#define pi(n) pf("%d\n",n)#define REP0(i,n) for(int i=0;i<(n);i++)#define REP1(i,n) for(int i=1;i<=(n);i++)#define REP(i,s,n) for(int i=s;i<=(n);i++)#define db double#define op operator#define pb push_back#define LL long long#define INF 0x3fffffff#define eps 1e-8#define PI acos(-1)#define maxn 1010using namespace std;int dp(int n){    if(n==0)rt 0;    if(n<=9)rt 1;    int tmp=1;    while(tmp<=n)tmp*=10;    tmp/=10;    int res;    if(n/tmp==1){        res=dp(tmp-1)+dp(n%tmp)+10*(n/10-tmp/10)+(n%10)+1;    }else {        res=(n/tmp)*dp(tmp-1)+dp(n%tmp)+tmp;    }    rt res;}int main(){    #ifdef ACBang    freopen("in.txt","r",stdin);    freopen("dp.txt","w",stdout);    #endif    int n;    while(~sf("%d",&n)){        int ans;        ans=dp(n);        pf("%d\n",ans);    }    rt 0;}