Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 482 | Accepted: 162 |
Description
Given a string of digits, insert commas to create a sequence of strictly increasing numbers so as to minimize the magnitude of the last number. For this problem, leading zeros are allowed in front of a number.
Input
Input will consist of multiple test cases. Each case will consist of one line, containing a string of digits of maximum length 80. A line consisting of a single 0 terminates input.
Output
For each instance, output the comma separated strictly increasing sequence, with no spaces between commas or numbers. If there are several such sequences, pick the one which has the largest first value;if there's a tie, the largest second number, etc.
Sample Input
Sample Output
Source
East Central North America 2002
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#include <stdio.h> #include <string.h> #define N_MAX 81 char seqStr[N_MAX]; char d[N_MAX][N_MAX]; char res[N_MAX],resTemp[N_MAX]; int num; bool Great(int start1,int len1,int start2,int len2) { while(len1 > 0 && seqStr[start1] == '0') { len1--; start1++; } while(len2 > 0 && seqStr[start2] == '0') { len2--; start2++; } if(len1 > len2) return true; else if(len1 == len2) { while(len1 > 0) { if(seqStr[start1] > seqStr[start2]) return true; else if(seqStr[start1] < seqStr[start2]) return false; else { start1++; start2++; len1--; } } return false; } else return false; } void GetResult(int idx,int len,int numTemp) { int i,j; if(idx == len) { if(num == 0) { resTemp[numTemp] = len; num = numTemp + 1; memcpy(res,resTemp,num); } else { bool best = false; resTemp[numTemp] = len; for(i = numTemp,j = num - 1 ; i >= 0 && j >= 0 ; i--,j--) { if(resTemp[i] > res[j]) best = true; if(resTemp[i] == res[j]) continue; break; } if(best == true) { num = numTemp + 1; memcpy(res,resTemp,num); } } } else { resTemp[numTemp] = len; for(i = 1 ; i <= d[idx][len] ; i++) { if(d[idx-len][i] == 0) continue; GetResult(idx-len,i,numTemp+1); } } } void DP() { int i,j,t,prev; bool bGreat; int len = strlen(seqStr + 1) + 1; d[1][1] = 1; for(i = 2 ; i < len ; i++) { d[i][i] = 1; for(j = 1 ; j < i ; j++) { prev = i - j; t = j; while(t < prev && seqStr[prev - t] == '0') t++; for( ; t > 0 ; t--) { if(d[prev][t] == 0) continue; if(Great(i - j + 1,j,prev - t + 1,t)) { bGreat = true; break; } } if(bGreat == true) d[i][j] = t; else d[i][j] = 0; } } prev = len - 1; for(i = 1 ; i < len ; i++) { if(d[prev][i] > 0) break; } j = i; while(j < len && seqStr[prev - j] == '0') j++; num = 0; for( ; i <= j ; i++) GetResult(prev,i,0); prev = 1; for(i = num - 1 ; i > 0 ; i--) { j = prev + res[i]; for( ; prev < j ; prev++) printf("%c",seqStr[prev]); printf(","); } j = prev + res[i]; for( ; prev < j ; prev++) printf("%c",seqStr[prev]); printf("/n"); } int main() { //freopen("test.txt","r",stdin); while(scanf("%s",seqStr + 1) != EOF) { if(strlen(seqStr + 1) == 1 && seqStr[1] == '0') break; DP(); } return 0; }
3,4,5,6 35,46 3,5,26 0001 100,000101
3456 3546 3526 0001 100000101 0