lc311309010103的专栏

#include#includeint main(){    int t,x,y;    char a;    scanf("%d",&t);      while(t--)      {         getchar();         scanf("%c %d %d",&a,&x,&y);         if(a=='+')        

#include#includeint main(){ int t,i,j,p; char s[100],a[100]; while(scanf("%d",&t)!=EOF) { getchar(); while(t--) { gets(s); p=strlen(s); j=0; for(i=str

#include#includeint main(){ int t,i; char s[60]; while(scanf("%d",&t)!=EOF) { getchar(); while(t--) { gets(s); for(i=0;i<strlen(s);i++)

QuicksumTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2196    Accepted Submission(s): 1676Problem DescriptionA checksum is an al

#include#includeint main(){ int i; char a[90]; while(gets(a)>0) { if(a[0]=='#') break; else { for(i=0;i<strlen(a);i++) { if(a

#includeint main(){    int k,t;    while(scanf("%d",&t)!=EOF)    {    while(t--)      {      scanf("%d",&k);           if(k=90)      printf("A\n");      else if(k=80)      prin

字符串统计Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 40761    Accepted Submission(s): 22714Problem Description对于给定的一个字符串，统计其中数字字符出现的

ParityTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2296    Accepted Submission(s): 1790Problem DescriptionA bit string has odd

字符串统计Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 40761    Accepted Submission(s): 22714Problem Description对于给定的一个字符串，统计其中数字字符出现的

#include#includeint main(){ int i,j,t; char a[60],b[60],c[60],d[60]; while(scanf("%d",&t)!=EOF) { getchar(); while(t--) { gets(a); gets

奇偶位互换Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4531    Accepted Submission(s): 3476Problem Description给定一个长度为偶数位的0,1字符串，请编程实

#include#includeint main(){    int i;    char s[200];    while(gets(s)!=NULL&&strcmp(s,"ENDOFINPUT")){     if(!strcmp(s,"START")||!strcmp(s,"END"))     continue;     for(i=0;i    

#includeint main(){ int t,i; char s[20]; while(scanf("%d",&t)!=EOF) { getchar(); while(t--) { gets(s); for(i=0;i<11;i++) { if(s[i]>='

ZOJTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1662    Accepted Submission(s): 1181Problem Description读入一个字符串，字符串中包含ZOJ三个字符，个数不一

ASCII码排序Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 85544    Accepted Submission(s): 35716Problem Description输入三个字符后，按各字符的ASCI

一、/*（1）怎样计数，存储，比较  答：只存储需要输出的出现最多的颜色的次数 ,进行一个动态变化  那么怎么使得颜色和次数对应呢 每次动态变化时不仅仅是次数 同时还有标记颜色的i */ #include&lt;stdio.h&gt;#include&lt;string.h&gt;#define N 1001int main(){    int T=1,n,s,max,i,j,x...