摘要：
Time时间限制：1000 ms | 内存限制：65535 KB 难度：2描述 Digital clock use 4 digits to express time, each digit is described by 3*3 characters (including”|”,”_”and” “).now given the current time, please tell us how can it be expressed by the digital clock.Hint:The digits showed by the digital clock are as follows: _ 阅读全文
摘要：
数的长度时间限制：3000 ms | 内存限制：65535 KB 难度：1描述 N！阶乘是一个非常大的数，大家都知道计算公式是N!=N*(N-1）······*2*1.现在你的任务是计算出N！的位数有多少（十进制）？输入 首行输入n，表示有多少组测试数据(n#include int main(){int n;scanf("%d",&n);while(n--){int m,i;double t,s=0;scanf("%d",&m);for(i=1;i<=m;i++){t=lo 阅读全文
摘要：
成绩转换Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 57281 Accepted Submission(s): 24928Problem Description输入一个百分制的成绩t，将其转换成对应的等级，具体转换规则如下：90~100为A;80~89为B;70~79为C;60~69为D;0~59为E;Input输入数据有多组，每组占一行，由一个整数组成。Output对于每组输入数据，输出一行。如果输入数据不在0~100范围内，请输出一行 阅读全文
摘要：
平方和与立方和Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 69792 Accepted Submission(s): 21706Problem Description给定一段连续的整数，求出他们中所有偶数的平方和以及所有奇数的立方和。Input输入数据包含多组测试实例，每组测试实例包含一行，由两个整数m和n组成。Output对于每组输入数据，输出一行，应包括两个整数x和y，分别表示该段连续的整数中所有偶数的平方和以及所有奇数的立方和。你可 阅读全文
摘要：
A+B for Input-Output Practice (III)Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 56513 Accepted Submission(s): 29030Problem DescriptionYour task is to Calculate a + b.InputInput contains multiple test cases. Each test case contains a pair of int 阅读全文
摘要：
A+B for Input-Output Practice (II)Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 47295 Accepted Submission(s): 31461Problem DescriptionYour task is to Calculate a + b.InputInput contains an integer N in the first line, and then N lines follow. Ea 阅读全文
摘要：
A+B for Input-Output Practice (I)Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 62478 Accepted Submission(s): 35144Problem DescriptionYour task is to Calculate a + b.Too easy?! Of course! I specially designed the problem for acm beginners. You mu 阅读全文