hdu 1015 Safecracker 枚举

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难度：0
算法：枚举
在一个字符串中赵武个字符v,w,x,y,z是的v - w^2 + x^3 - y^4 + z^5 = target ，其中v,w,x,y,z表示字符对应的数字,A对应1，B对应2，。。。，Z对应26。如果有多种情况，找字典树最大的。
数据量不是很大，所以枚举

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;int n , len , v , w , x , y , z;char ch[22];int get(char c) {return c - 'A' + 1;}int main() {while(~scanf("%d%s" , &n , ch)) {if(n == 0 && strcmp(ch , "END")==0) break;len = strlen(ch);char s1[22] = "00000" , s2[22];for(int i=0;i<len;i++) {for(int j=0;j<len;j++) {if(j == i) continue;for(int k=0;k<len;k++) {if(k == i || k == j) continue;for(int t=0;t<len;t++) {if(t == i || t == j || t == k) continue;for(int p=0;p<len;p++) {if(p == i || p == j || p == k || p == t) continue;v = get(ch[i]) , w = get(ch[j]) , x = get(ch[k]) , y = get(ch[t]) , z = get(ch[p]);if(v-w*w+x*x*x-y*y*y*y+z*z*z*z*z == n) {s2[0] = ch[i] , s2[1] = ch[j] , s2[2] = ch[k] , s2[3] = ch[t] , s2[4] = ch[p] , s2[5] = 0;if(strcmp(s1 , s2) < 0) strcpy(s1 , s2);}}}}}}if(s1[0] == '0') puts("no solution");else {printf("%s\n" , s1);}}return 0;}

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