12年校赛
0:
普通的字符串替换
代码:
#include<stdio.h>
#include<string.h>
char unit[9][5]={"I","II","IIII","IV","V","VI","VII","VIII","IX"};
char ten[9][5]={"X","XX","XXX","XL","L","LX","LXX","LXXX","XC"};
char hun[9][5]={"C","CC","CCC","CD","D","DC","DCC","DCCC","CM"};
int main(){
int n;
char num[10];
// for(int k=0;k<9;k++)
// printf("%s\n",ten[k]);
scanf("%d",&n);
while(n--){
scanf("%s",num);
int de;
int len=strlen(num);
de=len;
// printf("%d\n",len);
for(int i=0;i<len;i++,de--){
if(de==4) {
for(int j=0;j<num[i]-'0';j++){
printf("M");
}
}
else if(de==3){
printf("%s",hun[num[i]-'0'-1]);
}
else if(de==2){
// printf("%d %s",de,ten[8]);
printf("%s",ten[num[i]-'0'-1]);
// printf("%d\n",num[i]-'0'-1);
}
else if(de==1){
printf("%s",unit[num[i]-'0'-1]);
}
}
puts("");
}
return 0;
}
1:
二维数组的运用
代码:
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<vector>
using namespace std;
#define LL long long
#define STOP system("pause")
#define print(x) printf("%d\n",x);
#define eps 1e-8
#define PI acos(-1.0)
#define LMT 35
int mat[LMT][LMT];
int main(){
int i,j,T,im,ij,ii,n,m;
int yes;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
yes=0;
for(i=0;i<n;i++)
for(j=0;j<m;j++)
scanf("%d",&mat[i][j]);
for(i=0;i<n;i++)
{
im=mat[i][0];ij=0;
for(j=1;j<m;j++)
if(im<mat[i][j])
{
im=mat[i][j];
ij=j;
}
for(j=0;j<m;j++)
if(ij!=j&&mat[i][j]==im)
break;
if(j<m)continue;
for(ii=0;ii<n;ii++)
if(mat[ii][ij]>im)
break;
else if(mat[ii][ij]==im&&ii!=i)
break;
if(ii==n)
{
if(yes)printf(" ");
yes=1;
printf("%d",im);
}
}
if(0==yes)printf("NO\n");
else printf("\n");
}
return 0;
}
2:
约瑟夫环的利用，公式如下也可以用链模拟
代码：
#include<stdio.h>
int work(int n,int m)
{
int i=0,c;
for(c=1;c<=n;c++)
i=(i+m-1)%c+1;
return i;
}
int main(void)
{
int T,n;
scanf("%d",&T);
int k=T;
while(T--)
{
scanf("%d",&n);
printf("%d\n",work(n,3));
}
return 0;
}
3:
字符串的处理，比较麻烦,代码还没验证过，不过也差不多了。
#include<iostream>
#include<vector>
#include<string>
#include<stdio.h>
#include<string.h>
using namespace std;
#define N 5005
#define M 555
char str[N];
char words[M][M];
int x,y;
char aim[M][M];
void getwords(){
int len=strlen(str);
int i,j,k;
x=0;
y=0;
for(i=0;i<len;i++){
if(str[i]=='.'||(str[i]>='a'&&str[i]<='z')||(str[i]>='A'&&str[i]<='Z')){
words[x][y++]=str[i];
}
else {
words[x][y]='\0';
x++;
y=0;
}
}
}
void check(){
int i;
for(i=0;i<=x;i++)
printf("%s\n",words[i]);
}
void writeln(){
int a,b,i,j;
int l=0;
int num=0,k=0,ff=0;
for(a=0;a<=x;){
int len =strlen(words[a]);
// printf("%d\n",len);
if(l+len<=20){
memcpy(aim[k],words[a],strlen(words[a]));
// system("pause");
// printf("aim : %s words: %s\n",aim[k],words[a]);
a++;
k++;
l+=len;
num++;
}
else {
int space=20-l,nspace;
if(num==1) nspace=0;
else {
if(space%(num-1)!=0) ff=1;
nspace=space/(num-1);
}
for(i=0;i<k-1;i++){
printf("%s",aim[i]);
for(j=0;j<nspace;j++){
printf(" ");
}
}
if(ff==1) printf(" ");
ff==0;
printf("%s\n",aim[k-1]);
num=0;
l=0;
k=0;
memset(aim,0,sizeof(aim));
}
}
ff=0;
if(l!=0){
int space=20-l,nspace;
if(num==1) nspace=0;
else {
if(space%(num-1)!=0) ff=1;
nspace=space/(num-1);
}
for(i=0;i<k-1;i++){
printf("%s",aim[i]);
for(j=0;j<nspace;j++){
printf(" ");
}
}
// system("pause");
// printf("num: %d ff: %d k: %d",num,ff,k);
if(ff==1) printf(" ");
ff==0;
printf("%s\n",aim[k-1]);
num=0;
l=0;
k=0;
memset(aim,0,sizeof(aim));
}
}
int main(){
int T;
scanf("%d",&T);
getchar();
while(T--){
gets(str);
getwords();
// check();
writeln();
}
return 0;
}
4:dp.
代码:
#include<stdio.h>
#include<string.h>
#define LMT 25
double sum,rest[LMT],dis[LMT],cons[LMT],mon[LMT];
double v,ks,sd;
int main(void)
{
int T,i,j,d,n;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
scanf("%lf%lf%lf",&v,&rest[0],&ks);
for(i=0;i<=n+2;i++)
mon[i]=-9999;
for(i=1;i<=n;i++)
rest[i]=v;
mon[0]=0;
for(i=1;i<=n;i++)
{
scanf("%lf%lf",&dis[i],&cons[i]);
sum+=dis[i];
}
scanf("%lf",&dis[n+1]);
cons[n+1]=0;
for(i=1;i<=n+1;i++)
{
sd=dis[i];
for(j=i-1;j>=0;j--)
{
if(sd*ks<=rest[j])
{
if(mon[i]>mon[j]+(v-(rest[j]-sd*ks))*cons[i]||mon[i]<0)
mon[i]=mon[j]+(v-(rest[j]-sd*ks))*cons[i];
}
sd+=dis[j];
}
}
if(mon[n+1]>=0)printf("%.2lf\n",mon[n+1]);
else printf("NO\n");
}
return 0;
}
5:
已知后序中序求前序.
dfs
代码:
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<vector>
using namespace std;
#define LMT 35
#define LL long long
#define STOP system("pause")
#define print(x) printf("%d\n",x);
#define eps 1e-8
#define PI acos(-1.0)
char hou[LMT],mid[LMT],ans[LMT];
void dfs(int begin,int a,int b,int l,int r)
{
if(l>r||a>b)return;
int sign,i;
char x=hou[b];
for(i=l;i<=r&&x!=mid[i];i++);
sign=i;
ans[begin]=x;
dfs(begin+1,a,sign-l+a-1,l,sign-1);
dfs(begin+sign-l+1,sign-l+a,b-1,sign+1,r);
}
int main()
{
int T;
scanf("%d",&T);
while(scanf("%s%s",hou,mid)!=EOF)
{
memset(ans,0,sizeof(ans));
dfs(0,0,strlen(hou)-1,0,strlen(mid)-1);
printf("%s\n",ans);
}
return 0;
}
6:麻将考虑三种情况吧，实在是麻烦
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