POJ 3286- How many 0's?(组合数学_区间计数)
2015-04-22 15:33
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How many 0's?
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d
& %I64u
Submit Status Practice POJ
3286
Appoint description:
System Crawler (2015-04-18)
Description
A Benedict monk No.16 writes down the decimal representations of all natural numbers between and including m and n, m ≤ n. How many 0's will he write down?
Input
Input consists of a sequence of lines. Each line contains two unsigned 32-bit integers m and n, m ≤ n. The last line of input has the value of m negative and this line should not be processed.
Output
For each line of input print one line of output with one integer number giving the number of 0's written down by the monk.
Sample Input
Sample Output
题意:求区间内出现多少个零。和POJ 2282差不多。
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d
& %I64u
Submit Status Practice POJ
3286
Appoint description:
System Crawler (2015-04-18)
Description
A Benedict monk No.16 writes down the decimal representations of all natural numbers between and including m and n, m ≤ n. How many 0's will he write down?
Input
Input consists of a sequence of lines. Each line contains two unsigned 32-bit integers m and n, m ≤ n. The last line of input has the value of m negative and this line should not be processed.
Output
For each line of input print one line of output with one integer number giving the number of 0's written down by the monk.
Sample Input
10 11 100 200 0 500 1234567890 2345678901 0 4294967295 -1 -1
Sample Output
1 22 92 987654304 3825876150
题意:求区间内出现多少个零。和POJ 2282差不多。
#include <stdio.h> #include <math.h> #include <string.h> #include <stdlib.h> #include <iostream> #include <sstream> #include <algorithm> #include <set> #include <queue> #include <stack> #include <map> using namespace std; typedef long long LL; const int inf=0x3f3f3f3f; const double pi= acos(-1.0); #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 LL b[12]={1,10,100,1000,10000,100000,1000000,10000000,100000000,1000000000,10000000000,100000000000}; LL get_res(LL n) { int i; LL left,m; LL cnt=0; for(i=1;i<12;i++){ left=n/b[i]-1; cnt+=left*b[i-1]; m=(n%b[i]-n%b[i-1])/b[i-1]; if(m>0) cnt+=b[i-1]; else if(m==0) cnt+=n%b[i-1]+1; if(n<b[i]) break; } return cnt; } int main() { LL n,m; while(~scanf("%lld %lld",&n,&m)){ if(n==-1&&m==-1) break; if(n>m) swap(n,m); printf("%lld\n",get_res(m)-get_res(n-1)); } return 0; }
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