[CF]Sums of Digits
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挺有意思的一道题目,给你N个数字,其N个数是严格递增序列数字每个数位上的和。求出原来的递增序列。递增序列满足最小的情况。
#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <iostream>#include <cmath>#include <queue>#include <map>#include <stack>#include <list>#include <vector>#include <ctime>#define LL __int64#define eps 1e-8#define pi acos(-1)using namespace std;string ans[310];string go(int a,int b){string cnt="";int l=ans[a].length();for (int i=l-1;i>=0;i--){for (int j=ans[a][i]-'0'+1;j<=9;j++){cnt="";int s=j;for (int k=0;k<i;k++){s+=ans[a][k]-'0';cnt+=ans[a][k];}cnt+=char(j+'0');if (s+9*(l-1-i)>=b && s<=b){int g=b;for (int k=i+1;k<l;k++){for (int t=0;t<=9;t++)if (s+t+9*(l-k-1)>=g){cnt+=char(t+'0');g-=t;break;}}return cnt;}}}for (int i=l+1;;i++){for (int j=1;j<=9;j++){cnt="";int s=j;cnt=char(s+'0');if (s+9*(i-1)>=b && s<=b){int g=b;for (int k=0;k<i-1;k++){for (int t=0;t<=9;t++)if (s+t+9*(i-k-2)>=g){cnt+=char(t+'0');g-=t;break;}}return cnt;}}}}int main(){int n,k;scanf("%d",&n);//memset(ans,0,sizeof(ans));for (int i=1;i<=n;i++){scanf("%d",&k);ans[i]=go(i-1,k);}for (int i=1;i<=n;i++)cout<<ans[i]<<endl;return 0;}
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