UVa11038 How Many O's?
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1.题目描述:点击打开链接
2.解题思路:本题要求统计区间[m,n]中0出现的次数。可以利用加法原理,乘法原理来解决。不妨设solve(x)表示区间[0,x]中,0出现的次数。那么该如何计算这个函数呢?我们可以考虑逐位统计。设当前在第cur位,那么cur左边的数构成一个整数left,cur右边构成的整数为right,位数为k。根据乘法原理,如果第cur位不是0,那么left可以取[1,left]中任意一个数,right可以取10^k(因为每一位都可以取0~9,且是相互独立的),因此总的方案数就是left*10^k。而如果第cur位本身就是0,那么只有当左边取[0,left-1]的时候,右边可以有10^k种取法,而恰好为left时候,只能取[0,right],即此时的方案数为(left-1)*10^k+right+1。这样,累加每一位的方案数,就是答案。
3.代码:
#include<iostream>#include<algorithm>#include<cassert>#include<string>#include<sstream>#include<set>#include<bitset>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<cctype>#include<complex>#include<functional>#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define rep(i,n) for(int i=0;i<(n);i++)#define me(s) memset(s,0,sizeof(s))#define pb push_back#define lid (id<<1)#define rid (id<<1|1)typedef long long ll;typedef pair<int,int> P;ll solve(ll left){ if(left<0)return 0; ll ans=1,right=0,x=1; while(left>=10) { ll mid=left%10; left/=10; if(mid)ans+=left*x; else ans+=(left-1)*x+right+1; right+=mid*x; x*=10; } return ans;}int main(){ ll m,n; while(~scanf("%lld%lld",&m,&n)) { if(m==-1&&n==-1)break; ll L=solve(m-1); ll R=solve(n); printf("%lld\n",R-L); }}
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