日期:2014-05-20 浏览次数:20898 次
// 用java写的,
// 得出月天数为29天的时候, 才有=138的时候,
public class Test138 {
public static void main(String[] args) {
System.out.println(total(28, 7));
System.out.println(total(29, 7));
System.out.println(total(30, 7));
System.out.println(total(31, 7));
}
public static int total(int month, int count) {
int total = 0;
int s = 0;
for (int i = 1; i <= month; i++) {
s = i;
for (int j = 0; j < count; j++) {
if (s > month) {
s = 1;
}
total += s;
s++;
}
if (total == 138) {
total = i;
break;
} else {
total = 0;
}
}
return total;
}
}
------解决方案--------------------
第2题,假设7天都是同一个月那么设经过的第一天为X, 7项等差数列和138,X不能整除,所以应该跨月。
如果跨1天 , 和上面一样,6项等差数列和137 , X也是不能整除。
跨2天 , 5项等差数列和135 , X整除为25 , 那么这一天应该是24号了,而且这个月到29号就跨月了,所以应该是2月24号。
------解决方案--------------------
1
int sum = 20;
int bottle = sum;
while (bottle > 1) {
sum += bottle/2;
bottle = bottle%2 + bottle/2;
}
System.out.println(sum);
--结果 39
2
int[] end = {28, 29, 30, 31};
int days=138, count=0, day, max;
boolean found = false;
for (int i=0; i<end.length; i++) {
max = end[i] + 7;
day = 1;
while (max > 0) {
count = 0;
for (int j=day; j<day+7; j++) {
if (j > end[i]) {
count += j%end[i];
} else {
count += j;
}
}
if (count == days) {
found = true;
for (int j=day; j<day+7; j++) {
System.out.printf("%d, ", j>end[i] ? j%end[i] : j);
}
break;
}
day++;
max--;
}
if (found) {
System.out.printf("------end=%d\n", end[i]);
break;
}
}
if (!found) {
System.out.println("error");
}
------解决方案--------------------
1.20元能买20瓶水,然后20个瓶子可以换20/2瓶水,20/2个瓶子有可以换20/2/2瓶水,以此类推直到瓶子个数小于2为止。
public class T1{
public static void main(String[] args) {
int sum = 20;
int bottle = sum;
while (bottle > 1) {
sum += bottle/2;
bottle = bottle%2 + bottle/2;
}
System.out.println(sum);
}
}
2.7天加起来等于138,说明这个日期接近月末,所以要考虑这个月共有几天,最常见的是31天,30天,还有二月的28天和闰年二月的29天四个值。
public class T2 {
public static void main(String[] args) {
int[] end = {28, 29, 30, 31};
int days=138, count=0, day, max;
boolean found = false;
for (int i=0; i<end.length; i++) {
max = end[i] + 7;
day = 1;
while (max > 0) {
count = 0;
for (int j=day; j<day+7; j++) {
if (j > end[i]) {