codeforces#373B.Making Sequences is Fun
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题目链接:传送门
题意:给w,m,k,d(n)表示n这个数是几位数,w为总量,k为单个花费
从m开始如果的d(m)*k的花费小于w,则w减去这些花费,新的序列加入m
然后m+1,m+2…直到总量用完,求出序列的数的个数
做法就是求出m离10^(d(m))的花费能不能用完,能用完直接更新len,m更新到10^(d(m)),然后继续。
这题被套了,就是在判断花费的时候的乘法会爆longlong,所以判断里面要变成除法
代码如下:
#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;#define N 10010int d(long long int n){ int len=0; while(n){ n/=10; len++; } return len;}long long int Pow(int a,int b){ long long int r=1,base=a; while(b){ if(b&1) r*=base; base*=base; b/=2; } return r;}int main(){ long long int w,m; int k; cin>>w>>m>>k; int x=d(m); long long int len=0; while(w){ long long int y=Pow(10,x)-m; if(y<=(w/k)/x){ w-=(long long)y*k*x; len+=y; m=Pow(10,x); x=d(m); } else{ len+=(w/k)/x; w=0; } } cout<<len<<endl; return 0;}
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