The Little Elephant very much loves sums on intervals.
This time he has a pair of integers l and r (l?≤?r). The Little Elephant has to find the number of such integers x (l?≤?x?≤?r), that the first digit of integer x equals the last one (in decimal notation). For example, such numbers as 101, 477474 or 9 will be included in the answer and 47, 253 or 1020 will not.
Help him and count the number of described numbers x for a given pair l and r.
The single line contains a pair of integers l and r (1?≤?l?≤?r?≤?1018) — the boundaries of the interval.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64dspecifier.
On a single line print a single integer — the answer to the problem.
2 47
12
47 1024
98
In the first sample the answer includes integers 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44.
题意：给出两个数,n,m判断两个数间 满足头尾数字相同的数的个数
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>
#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)
#define eps 1e-8
#define pi acos(-1)
typedef __int64 ll;
#define fre(i,a,b) for(i = a; i <b; i++)
#define free(i,b,a) for(i = b; i >= a;i--)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define ssf(n) scanf("%s", n)
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define bug pf("Hi\n")
using namespace std;
#define N 100
ll a[N],b[20];
ll op[17];
void inint()
{
int i;
op[0]=1;
op[1]=10;
fre(i,2,18)
op[i]=op[i-1]*10;
b[0]=0;
b[1]=9;
b[2]=b[1]+9;
fre(i,3,19)
b[i]=b[i-1]+9*op[i-2]; //存下不大于长度为i个9的满足数目
}
ll hello(ll x)
{
ll i=0;
while(x)
{
a[i++]=x%10;
x/=10;
}
return i;
}
ll fdd(ll x)
{
ll i;
ll len=hello(x);
if(len==1) return x;
if(len==0) return 0;
ll le=a[len-1];
ll ri=a[0];
if(len==2) //防止 8548 中8538等情况
{
ll ans=0;
if(ri>=le) ans++;
le--;
if(ri) ans+=le;
ans+=9;
return ans;
}
ll temp=0;
//pf("%I64d %I64d %I64d\n",len,ri,le);
if(ri>=le) temp=1;
ri=le-1;
le=temp;
// pf("%I64d\n",le);
for(i=1;i<len-1;i++) //防止 8548 中8538等情况
{
le=le+a[i]*op[i-1];
//pf("%I64d %I64d %I64d %I64d\n",i,a[i],op[i-1],le);
}
// pf("%I64d\n",le);
if(ri>0)
le+=ri*op[len-2];
le=le+b[len-1]; //当前数目算完后加上位数少一个的数目
return le;
}
int main()
{
int i,j;
inint();
ll le,ri;
//freopen("in.txt","r",stdin);
while(~scanf("%I64d%I64d",&le,&ri))
{
le=fdd(le-1);
ri=fdd(ri);
// pf("%I64d %I64d\n",le,ri);
printf("%I64d\n",ri-le);
}
return 0;
}
/*
47 8545
2 47
47 1024
1000 1000
1 1000
47 74
*/
CF 205 C Little Elephant and Interval（模拟）
原文：http://blog.csdn.net/u014737310/article/details/44573079