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RSS订阅原创 A. Consecutive Factors
A. Consecutive Factors (20)时间限制400 ms内存限制65536 kB代码长度限制8000 B判题程序Standard作者CHEN, YueAmong all the factors of a positive intege
2015-03-29 20:55:33
687
原创 PAT A1010 Radix （有点问题）
1010. Radix (25)时间限制400 ms内存限制65536 kB代码长度限制16000 B判题程序Standard作者CHEN, YueGiven a pair of positive integers, for example, 6 an
2015-03-09 21:20:06
421
原创 PAT A1009. Product of Polynomials
1009. Product of Polynomials (25)时间限制400 ms内存限制65536 kB代码长度限制16000 B判题程序Standard作者CHEN, YueThis time, you are supposed to find
2015-03-08 21:59:04
343
原创 PAT A1007Maximum Subsequence Sum（贪心算法）
#include int main() {int num[10000];int path[10000][3];int K, i, temp, start, end, max, max_start, max_end;scanf("%d", &K);for(i = 0; i scanf("%d", &num[i]);}for(i = 0; i if(i
2015-03-08 20:49:00
379
原创 PAT A 1007. Maximum Subsequence Sum(贪心算法应用path[i] 与 path[i + 1)
1007. Maximum Subsequence Sum (25)时间限制400 ms内存限制65536 kB代码长度限制16000 B判题程序Standard作者CHEN, YueGiven a sequence of K integers { N
2015-03-08 20:40:37
426
原创 PAT A 1006. Sign In and Sign Out (25)
1006. Sign In and Sign Out (25)时间限制400 ms内存限制65536 kB代码长度限制16000 B判题程序Standard作者CHEN, YueAt the beginning of every day, the fi
2015-03-08 15:37:56
356
原创 PAT A1002. A+B for Polynomials
1002. A+B for Polynomials (25)时间限制400 ms内存限制65536 kB代码长度限制16000 B判题程序Standard作者CHEN, YueThis time, you are supposed to find A+
2015-03-08 10:26:46
373
原创 PAT A1035. Password (20)
1035. Password (20)时间限制400 ms内存限制65536 kB代码长度限制16000 B判题程序Standard作者CHEN, YueTo prepare for PAT, the judge sometimes has to ge
2015-03-07 22:03:27
308
原创 PAT A1031. Hello World for U
1031. Hello World for U (20)时间限制400 ms内存限制65536 kB代码长度限制16000 B判题程序Standard作者CHEN, YueGiven any string of N (>=5) characters,
2015-03-07 21:20:32
304
原创 PAT A 1027. Colors in Mars
1027. Colors in Mars (20)时间限制400 ms内存限制65536 kB代码长度限制16000 B判题程序Standard作者CHEN, YuePeople in Mars represent the colors in thei
2015-03-07 20:35:48
315
原创 PAT A1023. Have Fun with Numbers
1023. Have Fun with Numbers (20)时间限制400 ms内存限制65536 kB代码长度限制16000 B判题程序Standard作者CHEN, YueNotice that the number 123456789 is
2015-03-07 15:10:13
342
原创 PAT A1015. Reversible Primes (20)
1015. Reversible Primes (20)时间限制400 ms内存限制65536 kB代码长度限制16000 B判题程序Standard作者CHEN, YueA reversible prime in any number system
2015-03-07 14:28:28
249
原创 PAT A1019. General Palindromic Number
1019. General Palindromic Number (20)时间限制400 ms内存限制65536 kB代码长度限制16000 B判题程序Standard作者CHEN, YueA number that will be the same
2015-03-07 14:27:03
236
原创 PAT A1015. Reversible Primes
1015. Reversible Primes (20)时间限制400 ms内存限制65536 kB代码长度限制16000 B判题程序Standard作者CHEN, YueA reversible prime in any number system
2015-03-07 13:33:53
320
原创 PAT A1015. Reversible Primes
1015. Reversible Primes (20)时间限制400 ms内存限制65536 kB代码长度限制16000 B判题程序Standard作者CHEN, YueA reversible prime in any number system
2015-03-07 13:32:53
376
原创 PAT A1015. Reversible Primes
1015. Reversible Primes (20)时间限制400 ms内存限制65536 kB代码长度限制16000 B判题程序Standard作者CHEN, YueA reversible prime in any number system
2015-03-07 13:25:41
299
原创 PAT A1011. World Cup Betting
1011. World Cup Betting (20)时间限制400 ms内存限制65536 kB代码长度限制16000 B判题程序Standard作者CHEN, YueWith the 2010 FIFA World Cup running, fo
2015-03-07 10:55:48
305
原创 PAT A1008. Elevator
1008. Elevator (20)时间限制400 ms内存限制65536 kB代码长度限制16000 B判题程序Standard作者CHEN, YueThe highest building in our city has only one ele
2015-03-07 10:18:20
368
原创 PAT A 1005. Spell It Right
Given a non-negative integer N, your task is to compute the sum of all the digits of N, and output every digit of the sum in English.Input Specification:Each input file contains one test case.
2015-03-06 23:45:28
285
原创 PAT A1001. A+B Format
1001. A+B Format (20)时间限制400 ms内存限制65536 kB代码长度限制16000 B判题程序Standard作者CHEN, YueCalculate a + b and output the sum in standard
2015-03-06 23:30:45
380
原创 PAT 1024. 科学计数法
1024. 科学计数法 (20)时间限制100 ms内存限制65536 kB代码长度限制8000 B判题程序Standard作者HOU, Qiming科学计数法是科学家用来表示很大或很小的数字的一种方便的方法，其满足正则表达式[+-][1-9]"."[
2015-03-06 22:33:43
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原创 PAT 1023. 组个最小数
1023. 组个最小数 (20)时间限制100 ms内存限制65536 kB代码长度限制8000 B判题程序Standard作者CAO, Peng给定数字0-9各若干个。你可以以任意顺序排列这些数字，但必须全部使用。目标是使得最后得到的数尽可能小（注意
2015-03-06 19:37:52
934
原创 PAT 1022. D进制的A+B
1022. D进制的A+B (20)时间限制100 ms内存限制65536 kB代码长度限制8000 B判题程序Standard作者CHEN, Yue输入两个非负10进制整数A和B(30-1)，输出A+B的D (1 输入格式：输入在一行
2015-03-06 17:46:49
1586
原创 PAT 1020. 月饼
#include int main() {int N, D, i, j, max_i;float sum, remain, max;float num[4][1000];scanf("%d %d", &N, &D);for(i = 0; i for(j = 0; j scanf("%f", &num[i][j]);}}for(i = 0; i
2015-03-06 16:22:56
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原创 PAT 1019. 数字黑洞
#include void sortNum(int* a, int* b, int* c, int* d); int main() {
2015-03-06 10:05:36
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原创 PAT 1018. 锤子剪刀布
#include void printMax(int a, int b, int c) {if(a >= b && a >= c) {printf("B");} else if(b > a && b >= c) {printf("C");} else if(c > a && c > b) {printf("J");}}int main() {int
2015-03-05 22:17:59
453
原创 PAT 1014. 福尔摩斯的约会
#include int main() {char str1[70], str2[70], str3[70], str4[70];char week[7][5] = {"MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN"};char *p, *q;int flag, i;scanf("%s", str1);scanf("
2015-03-05 20:05:55
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原创 PAT 1013. 数素数
#include #include int prime(int num) {int sqr, flag, i;sqr = (int) sqrt(num);flag = 1;for(i = 2; i if(num % i == 0) {flag = 0;break;}} return flag;}int main() {int M,
2015-03-03 23:15:11
595
原创 PAT 1017. A除以B
#include #include int main() {char A[1100], Q[1100];int B, R, i, j, length, sum;scanf("%s", A);scanf("%d", &B);length = strlen(A);sum = 0;for(i = 0, j = 0; i sum = 10 * sum + A
2015-03-03 22:40:13
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原创 PAT 1009. 说反话
#include #include int main() {char str[90];char *p, *s;int length;gets(str);length = strlen(str);for(p = str + length - 1; p != str; p --) {if(*p == ' ') {for(s = p + 1; *s !
2015-03-03 19:39:21
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原创 1007. 素数对猜想
#include #include int prime(int num) {int i, flag;flag = 1;for(i = 2; i if(num % i == 0) {flag = 0;break;}}return flag;}int main() {int i, n, num;scanf("%d", &n);fo
2015-03-03 16:23:53
412
原创 PAT 1006. 换个格式输出整数
#include int main() {int n, a, b, c, i;scanf("%d", &n);a = n / 100;b = (n % 100) / 10;c = n % 10;if(a > 0) {for(i = 0; i printf("B");}}if(b > 0) {for(i = 0; i printf("S
2015-03-03 15:54:36
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原创 PAT 1005. 继续(3n+1)猜想
#include int main() {int num[101];int i, k, n, length; for(i = 0; i num[i] = -1;}scanf("%d", &k);for(i = 0; i scanf("%d", &n);num[n] = 0;}for(i = 0; i if(num[i] != 0) {
2015-03-03 10:26:26
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原创 PAT B1003 我要通过
#include #include int main() {char str[110];int n, length, flag, leftNum, midNum, rightNum, i, j;scanf("%d", &n);for(i = 0; i scanf("%s", str);length = strlen(str);flag = 1;leftNum
2015-03-03 09:05:43
1080
原创 PAT B1004 成绩排名
#include #include int strGrade(char str[]) {int length;length = strlen(str);if(str[length - 2] == ' ') {return str[length - 1] - '0';} else if(str[length -3] == ' ') {return 10 * (
2015-03-03 09:01:47
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