lc311309010103的专栏

数列有序!Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 45490    Accepted Submission(s): 19723Problem Description有n(n  Input

成绩转换Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 74034    Accepted Submission(s): 32538Problem Description输入一个百分制的成绩t，将其转换成对应的等级，

小明A+BTime Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 25242    Accepted Submission(s): 11985Problem Description小明今年3岁了, 现在他已经能够认识100以

#include#includeint fun(int p){     int n=0;    while(p>9)    {              n+=p%10;              p=p/10;              }              n+=p;              return n;    }int ma

#includeint main(){   int n,k,sum=0;   while(scanf("%d",&n)&&n!=0)   {        while(n--)        {        scanf("%d",&k);        sum+=k/100;        {        if((k%100)>=50)     

#includeint main(){    long int S,M,seed;    int i,n;    while(scanf("%ld%ld",&S,&M)!=EOF)    {    n=0;    seed=0;    for(i=1;;i++)    {    seed=(seed+S)%M;    if((seed==(0+S)%

蟠桃记Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 34389    Accepted Submission(s): 25896Problem Description喜欢西游记的同学肯定都知道悟空偷吃蟠桃的故事，你

#includeint main(){ int t,i,s,n; while(scanf("%d",&t)!=EOF) { while(t--) { s=0; scanf("%d",&n); while(n>0) { if(n%2==1)

#includeint main(){ int i,n,m,p,t; int a[10]; while(scanf("%d",&n)!=EOF) { i=0; m=n; while(n>0) { a[i++]=n%10; n/=10;} p=i; t=1; for(i=0;i<p;i++)

#includeint main(){    int n,m;    while(scanf("%x%x",&m,&n)!=EOF)    {     printf("%d\n",m+n);                                   }    return 0;    }

#include#includeusing namespace std;int main(){ int t,a[5],k,i; while(scanf("%d",&t)!=EOF) { k=0; while(t--) { for(i=0;i<3;i++) scanf("%d",&a[i]);

#includeint main(){ int m,s,k,i,p; while(scanf("%d%d",&m,&k)!=EOF&&(m!=0||k!=0)) { s=m; while(m>=k) { s+=m/k; m=m%k+m/k;

Build a FenceTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 943    Accepted Submission(s): 744Problem DescriptionThere is a

find the nth digitTime Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8385    Accepted Submission(s): 2394Problem Description假设：S1 = 1S2

#include#includeint main(){ int n; while(scanf("%d",&n)!=EOF) { printf("%d\n",(abs(n-1)+1)*(n+1)/2); }return 0; }

新生晚会Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8306    Accepted Submission(s): 2947Problem Description开学了，杭电又迎来了好多新生。ACMer想为新

#includeint main(){       int m,n,i,t;    long int x,y;    while(scanf("%d%d",&m,&n)!=EOF)    {         i=m+n;       m=m       n=i-m;        x=0,y=0;       for(i=m;i       { 

#includeint main(){    int m,n;    while(scanf("%d%d",&m,&n)!=EOF&&(m!=0||n!=0))    {    int r=0,p=0;    while(m>0)    {    r=r*10+m%10;    m/=10;} while(n>0)    {    p=p*1

Secret NumberTime Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7808    Accepted Submission(s): 3327Problem Description有一天, KIKI 收到一张

三角形Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 47193    Accepted Submission(s): 15891Problem Description给定三条边，请你判断一下能不能组成一个三角形

#includeint main(){    int n,u,d,s,m;        while(scanf("%d%d%d",&n,&u,&d)!=EOF&&(n!=0||u!=0||d!=0))    {      s=0;      m=0;      while(1)      {      s+=u;      m++;     

#includeint main(){ int t,m; char a; while(scanf("%d",&t)!=EOF) { getchar(); while(t--) { scanf("%c%d",&a,&m); if(a<'a') printf("%d\n",a+m-64); else p

奇偶数分离时间限制：3000 ms  |  内存限制：65535 KB难度：1描述有一个整型偶数n(2<= n <=10000),你要做的是：先把1到n中的所有奇数从小到大输出，再把所有的偶数从小到大输出。输入第一行有一个整数i（2每组有一个整型偶数n。输出第一行输出所有的奇数第二行输出所有的偶数样例输入21014样例

A hard puzzleTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 29044    Accepted Submission(s): 10408Problem Descriptionlcy gives a

三位数逆序输出时间限制：1000 ms  |  内存限制：65535 KB难度：1描述输入一个三位正整数，将其三个数位逆序颠倒后输出。输入有多组测试数据，以EOF结束。每组数据输入一个三位正整数。输出每组数据输出一行。样例输入789样例输出987来源2012年C语言上机题上传者A

#includeint main(){    int t,i,n,m,a,b;    int k,c=0;    while(scanf("%d",&t)!=EOF)    {     for(i=0;i      {                     k=1;        while(scanf("%d%d",&n,&m)&&(m!=0||n!=0

#includeint main(){    int n,m,s,t=0;    while(scanf("%d",&n)!=EOF&&(n>0))    {     s=0;      while(n--)      {       scanf("%d",&m);       s+=m;                }     printf("S

#includeint main(){    int n;    int p=0;    while(scanf("%d",&n)!=EOF&&n!=0)    {      p=0;      if(n==0)      printf("%d\n",p);      while(n!=1)      {      if(n%2==0)     

#includeint main(){      int n,m,i,t,l,a,b;      while(scanf("%d%d",&n,&m)!=EOF)      {      t=n/m;      l=n%m;      if(l==0)      {      for(i=1;i      {        a=(2*i-1)*m+1;

不要62Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 19032    Accepted Submission(s): 6435Problem Description杭州人称那些傻乎乎粘嗒嗒的人为62（音：la

#includeint main(){    int m,n;    int a[110];    while(scanf("%d%d",&m,&n)!=EOF&&(m!=0||n!=0))    {              int i,t=0;      for(i=0;i      {      if((m*100+i)%n==0)     

#includeint main(){    int a,b,k;    while(scanf("%d%d",&a,&b)!=EOF&&(a!=0||b!=0))    {    int i,m,n;    int j=0;    m=a;    n=b;    scanf("%d",&k);    for(i=0;i    {      

#include&lt;stdio.h&gt;//若用递归函数实现应该怎么操作 int main(){    int a,b,n,i,d,k;    int s[51]={0,1,1};//定义长度为51，s[0],s[1],s[2]分别为0,1,1的数组     while(scanf("%d%d%d",&amp;a,&amp;b,&amp;n)!=EOF&amp;&amp;a+b+n...