标签：cond mit number with ++ content should include next
InputFirst line contains an integer TT, which indicates the number of test cases.
Every test case contains one line with two integers xx, KK.
Limits
1≤T≤1001≤T≤100
0≤x≤1090≤x≤109
1≤K≤91≤K≤9OutputFor every test case, you should output ‘Case #x: y‘, where x indicates the case number and counts from 1 and y is the result.Sample Input
2 2 2 3 2
Sample Output
Case #1: 1 Case #2: 2
题意：给定函数f(y,K)=∑y的所有位的K次幂，给出X，K，问多少个Y满足f(Y,K)=X；
思路：似乎没有什么思路，考虑暴力，估计Y只有10位，所以我们暴力前面5位，用map去寻找后面5位。
然而我开始些了离散化WA了，map过了。
#include<bits/stdc++.h> #define ll long long #define rep(i,a,b) for(int i=a;i<=b;i++) using namespace std; const int maxn=1000010; unordered_map<ll,int>mp; ll tot,cnt,f[10][10],K,a[maxn],X,ans,num[maxn]; int main() { int T,Cas=0; scanf("%lld",&T); rep(i,0,9) f[i][0]=1; rep(i,1,9) rep(j,1,9) f[i][j]=f[i][j-1]*i; while(T--){ tot=cnt=0; mp.clear(); scanf("%lld%lld",&X,&K); ans=0; rep(i,0,9) rep(j,0,9) rep(k,0,9) rep(p,0,9) rep(q,0,9){ ll t1=f[i][K]+f[j][K]+f[k][K]+f[p][K]+f[q][K],t2=i*10000+j*1000+k*100+p*10+q; a[++tot]=t1-100000LL*t2; mp[t1-t2]++; if(t1-t2==X) ans++; } rep(i,2,tot) if(mp.find(X-a[i])!=mp.end())ans+=mp[X-a[i]]; printf("Case #%d: %lld\n",++Cas,ans-(X==0)); } return 0; }
错误代码（依然没有找到bug）：
#include<bits/stdc++.h> #define ll long long #define rep(i,a,b) for(int i=a;i<=b;i++) using namespace std; const int maxn=2000010; int Laxt[maxn],Next[maxn],tot,cnt,f[20][20],num[maxn]; ll a[maxn],To[maxn],X,ans,K; int find(ll x){ ll t=(x%maxn+maxn)%maxn; for(int i=Laxt[t];i;i=Next[i]) if(To[i]==x) return num[i]; return 0; } void add(ll x){ if(x==X) ans++; ll t=(x%maxn+maxn)%maxn; for(int i=Laxt[t];i;i=Next[i]){ if(To[i]==x) { num[i]++; return ;} } Next[++cnt]=Laxt[t]; Laxt[t]=++cnt; To[cnt]=x; num[cnt]=1; } int main() { int T,Cas=0; scanf("%lld",&T); rep(i,0,9) f[i][0]=1; rep(i,1,9) rep(j,1,9) f[i][j]=f[i][j-1]*i; while(T--){ memset(Laxt,0,sizeof(Laxt)); tot=cnt=0; scanf("%lld%lld",&X,&K); ans=0; rep(i,0,9) rep(j,0,9) rep(k,0,9) rep(p,0,9) rep(q,0,9){ ll t1=f[i][K]+f[j][K]+f[k][K]+f[p][K]+f[q][K],t2=i*10000+j*1000+k*100+p*10+q; a[++tot]=t1-100000LL*t2; add(t1-t2); } rep(i,2,tot) ans+=find(X-a[i]); printf("Case #%d: %lld\n",++Cas,ans-(X==0)); } return 0; }
标签：cond mit number with ++ content should include next
原文地址：https://www.cnblogs.com/hua-dong/p/9800794.html
