BestCoder 2nd Anniversary - 1001 Oracle
BestCoder 2nd Anniversary - 1001 OracleAccepts: 631 Submissions: 2576 Problem Description The youngest and most beautiful is Psyche,whose admirers,neglecting the proper worship of the love goddess Venus,instead pray and make offerings to her. Her father,the king,is desperate to know about her destiny,so he comes to the Delphi Temple to ask for an oracle. The oracle is an integer n n without leading zeroes. To get the meaning,he needs to rearrange the digits and split the number into two positive integers without leading zeroes,and their sum should be as large as possible. Help him to work out the maximum sum. It might be impossible to do that. If so,print Uncertain. Input For each test case,the single line contains an integer n n (1 le n < 10 ^ {10000000}) (1≤n<10 Output Sample Input Sample Output Hint In the first example,it is optimal to split 112 112 into 21 21 and 1 1,and their sum is 21 + 1 = 22 21+1=22. In the second example,it is optimal to split 233 233 into 2 2 and 33 33,and their sum is 2 + 33 = 35 2+33=35. In the third example,it is impossible to split single digit 1 1 into two parts. 题意:把一串数分为两部分正数,要求和值最大,否则,输出”Uncertain” #include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
using namespace std;
const int maxn = 10000100;
int n,m,k;
char s[maxn],a[maxn];
int main(){
scanf("%d",&n);
while(n--){
memset(a,' ',sizeof(a));
scanf("%s",s);
int len = strlen(s);
if(len==1){//一位数
printf("Uncertainn");
continue;
}
if(len==2 ){//两位数
if(s[0]=='0'||s[1]=='0') printf("Uncertainn");
else printf("%dn",s[0]-'0'+s[1]-'0');
continue;
}
sort(s,s+len);//升序
k = 0;
int flag = 1;
int si = -1;
int cnt0 = 0;
for(int i=0;i<len;i++){
if(s[i]!='0' && flag){
si = i;//第一次不为0
flag = 0;
continue;
}
if(s[i]=='0')cnt0++;
a[k++] = s[i];
}
if(cnt0>=len-1) {//只有一个不为0的数
printf("Uncertainn");
continue;
}
a[0] += s[si]-'0';
if(a[0] > '9'){
a[0] -= 10;
a[1] += 1;
}
m = k;
for(int i=1;i<k;i++){
if(i==k-1&&a[i]>'9') {
m++;
a[i] -= 10;
a[i+1] = '1';
continue;
}
if(a[i] > '9'){
a[i] -= 10;
a[i+1] += 1;
}
}
for(int i=m-1;i>=0;i--){
printf("%c",a[i]);
}
printf("n");
}
return 0;
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