Codeforces Round #388 (Div. 2)A Bachgold Problem
2017-01-23 15:57
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题目大意:
把一个大于1的数尽可能多的分成多个质数。
题目解法:
本质就是将偶数都分成2,质数分成一个3和多个2。
代码:
#include "iostream"
#include "cstdio"
#include "math.h"
#include "algorithm"
#include "string"
#include "string.h"
#include "vector"
#include "map"
#include "queue"
#include "bitset"
using namespace std;
int n;
int main() {
while (scanf("%d", &n) != EOF) {
if (n % 2) {
int num = (n - 3) / 2;
printf("%d\n", num + 1);
for (int i = 0;i < num;i++)
printf("2 ");
printf("3 ");
}
else {
printf("%d\n", n / 2);
for (int i = 0;i < n / 2;i++)
printf("2 ");
}
puts("");
}
return 0;
}
把一个大于1的数尽可能多的分成多个质数。
题目解法:
本质就是将偶数都分成2,质数分成一个3和多个2。
代码:
#include "iostream"
#include "cstdio"
#include "math.h"
#include "algorithm"
#include "string"
#include "string.h"
#include "vector"
#include "map"
#include "queue"
#include "bitset"
using namespace std;
int n;
int main() {
while (scanf("%d", &n) != EOF) {
if (n % 2) {
int num = (n - 3) / 2;
printf("%d\n", num + 1);
for (int i = 0;i < num;i++)
printf("2 ");
printf("3 ");
}
else {
printf("%d\n", n / 2);
for (int i = 0;i < n / 2;i++)
printf("2 ");
}
puts("");
}
return 0;
}
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