HDOJ1015看懂題之後的簡單粗暴

標籤：acm   hdoj   

/*看懂題意之後，給定target和大寫字串，即是：把ABCDE……轉換為12345…… 在給定的不重複數之中找5個數，使得其滿足a-b^2+c^3-d^4+e^5等於給定的數target由於資料量不大，最大為20個不重複大寫字母，不多說，5重for迴圈搞定*/#include <iostream>#include <algorithm>#include <stdio.h>#include <math.h>#include <map>#include <set>#include <vector>#include <string>#include <cstring>#include <sstream>using namespace std;#define input freopen("input.txt","r",stdin);#define output freopen("output.txt","w",stdout);#define For1(i,a,b) for (i=a;i<b;i++)#define For2(i,a,b) for (i=a;i<=b;i++)#define Dec(i,a,b) for (i=a;i>b;i--)#define Dec2(i,a,b) for (i=a;i>=b;i--)#define Sca_d(x) scanf("%d",&x)#define Sca_s(x) scanf("%s",x)#define Sca_c(x) scanf("%c",&x)#define Sca_f(x) scanf("%f",&x)#define Sca_lf(x) scanf("%lf",&x)#define Fill(x,a) memset(x,a,sizeof(x))#define MAXN 0x7fffffffint a,b,c,d,e;int book[30];int ans[10];int main(){int x,i,j,k,l;char ch[20];while(cin>>x>>ch){if (!x) break;Fill(book,0);Fill(ans,0);l=strlen(ch);For1(i,0,l) book[ch[i]-64]++;//數字字元統計 For2(a,1,26)if (book[a])For2(b,1,26)if (a!=b&&book[b])For2(c,1,26)if (c!=a&&c!=b&&book[c])For2(d,1,26)if (d!=a&&d!=b&&d!=c&&book[d])For2(e,1,26)if (e!=a&&e!=b&&e!=c&&e!=d&&book[e])if (a-b*b+c*c*c-d*d*d*d+e*e*e*e*e==x)ans[1]=a,ans[2]=b,ans[3]=c,ans[4]=d,ans[5]=e;//直接暴力到最後一組 if (ans[1]==0)//如果直到搜尋完畢還沒有解的話，nocout<<"no solution";elseFor2(i,1,5)printf("%c",ans[i]+64);//記得將其轉換為字元輸出cout<<endl;}return 0;}

HDOJ1015看懂題之後的簡單粗暴