要求和程序如下,不知道错在哪里,
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要求和程序如下,不知道错在哪里,
Uncle Da is very smart.His IQ was just 0 on his birthday.But it grows 1 everyday.For example,
when Uncle Da is 1 year old,his IQ is 365!
Ziyao big god is smarter then Uncle Da.His IQ was 1 on his birthday,and grows 1 time every day.
For example,when Ziyao is 1 year old,his IQ is 2^365!
Uncle Da's IQ is x when he is n years old.Ziyao's IQ is y when he is m years old.
(We assume that every year is 365 days)
Now a martian want to know how much is (y-x)%400009.
【input format】
Two numbers,n and m.(x,y
Uncle Da is very smart.His IQ was just 0 on his birthday.But it grows 1 everyday.For example,
when Uncle Da is 1 year old,his IQ is 365!
Ziyao big god is smarter then Uncle Da.His IQ was 1 on his birthday,and grows 1 time every day.
For example,when Ziyao is 1 year old,his IQ is 2^365!
Uncle Da's IQ is x when he is n years old.Ziyao's IQ is y when he is m years old.
(We assume that every year is 365 days)
Now a martian want to know how much is (y-x)%400009.
【input format】
Two numbers,n and m.(x,y
![要求和程序如下,不知道错在哪里,](/uploads/image/z/3544378-34-8.jpg?t=%E8%A6%81%E6%B1%82%E5%92%8C%E7%A8%8B%E5%BA%8F%E5%A6%82%E4%B8%8B%2C%E4%B8%8D%E7%9F%A5%E9%81%93%E9%94%99%E5%9C%A8%E5%93%AA%E9%87%8C%2C)
虽然不知道哪里错了,但我这有个运行正确的.
int n, m, i, j, sum = 0;
scanf("%d%d", &n, &m);
int x = (365 * n) % 400009 , y = 1;
for (j = 1; j 400009) y -= 400009;
}
printf("%d\n", (y - x + 400009) % 400009);
int n, m, i, j, sum = 0;
scanf("%d%d", &n, &m);
int x = (365 * n) % 400009 , y = 1;
for (j = 1; j 400009) y -= 400009;
}
printf("%d\n", (y - x + 400009) % 400009);
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