有时一些Codeforces上的implementation题也挺有意思的(给我等菜鸟继续刷题的信心啊!!)
codeforces 420 A. Start Up
http://codeforces.com/problemset/problem/420/A
大意:给一字符串,求解是否是镜像串。
(读懂题了就行)
镜像文字: 逆序看和原来是一样的。所以对单个字母也是有要求的,比如S wrong,A right。
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N=1e5+10;
char s[N];
char str[15]="AHIMOTUVWXY";
bool check(int len){
int left=0,right=len-1;
while(left<=right){
if(s[left]!=s[right]){
return 0;
}
left++;
right--;
}
return 1;
}
int main()
{
//freopen("cin.txt","r",stdin);
//str="AHIMOTUVWXY";
while(~scanf("%s",s)){
int len=strlen(s);
bool ans=0;
if(check(len)){
ans=1;
for(int i=0;i<len;i++){
bool tag=0;
for(int j=0;j<11;j++){
if(s[i]==str[j]){
tag=true;
break;
}
}
if(tag==0){
ans=0;
break;
}
}
}
if(ans) puts("YES");
else puts("NO");
}
return 0;
}
codeforces 417 A. Elimination
http://codeforces.com/problemset/problem/417/A
读懂题后,建立方程,然后暴力解方程:
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int c,d,n,m,k;
while(~scanf("%d%d%d%d%d",&c,&d,&n,&m,&k)){
int len1=m,len2=n*m,ans=1<<29; //直接暴力解方程
for(int x=0;x<=len1;x++){
for(int y=0;y<=len2;y++){
if(n*x+y+k>=len2){
ans=min(ans,c*x+d*y);
}
}
}
printf("%d\n",ans);
}
return 0;
}
codeforces 417B - Crash
http://codeforces.com/problemset/problem/417/B
把握好题意:时间顺序。chronological order
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N=1e5+10;
int f[N];
int main()
{
//freopen("cin.txt","r",stdin);
int n,x,k;
while(cin>>n){
memset(f,-1,sizeof(f));
bool flag=1;
for(int i=0;i<n;i++){
scanf("%d%d",&x,&k);
if(x==f[k]+1) f[k]++;
else if(x>f[k]+1) flag=0;
}
if(flag) puts("YES");
else puts("NO");
}
return 0;
}
codeforces 418 A. Football
http://codeforces.com/problemset/problem/418/A
矩阵记录相通的信息。(邻接矩阵的思想)
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
bool map[1005][1005];
int main()
{
int n,k;
while(~scanf("%d%d",&n,&k)){
memset(map,0,sizeof(map));
bool flag=1;
for(int i=1;i<=n;i++){
for(int j=i+1,d=0;d<k;d++,j++){
j=(j%n==0?n:j%n);
map[i][j]=1;
if(map[j][i]==1){
flag=0;
break;
}
}
if(!flag) break;
}
if(flag){
printf("%d\n",n*k);
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
if(map[i][j]) printf("%d %d\n",i,j);
}
}
}
else printf("-1\n");
}
return 0;
}
codeforces 413 A. Data Recovery
http://codeforces.com/problemset/problem/413/A
If the data is consistent 是否合理,是否一致
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int a[110];
int main()
{
//freopen("cin.txt","r",stdin);
int n,m,maxm,minm;
while(~scanf("%d%d%d%d",&n,&m,&minm,&maxm)){
for(int i=1;i<=m;i++) scanf("%d",&a[i]);
sort(a+1,a+m+1);
a[0]=minm;
a[m+1]=maxm;
int sum=m;
if(a[m+1]>a[m]) sum++;
if(a[0]<a[1]) sum++;
if(a[m+1]<a[m]||a[0]>a[1]) sum=-1;
if(sum<0||sum>n) puts("Incorrect"); //
else puts("Correct");
}
return 0;
}
codeforces 413 B. Spyke Chatting
http://codeforces.com/problemset/problem/413/B
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long LL;
const int N=2e4+10;
int g[N][12]; //employee接听
LL h[12]; //chat接收
LL f[N]; // employee 发出
LL fa[N][12];
int main()
{
//freopen("cin.txt","r",stdin);
int n,m,k;
while(~scanf("%d%d%d",&n,&m,&k)){
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
scanf("%d",&g[i][j]);
}
}
memset(h,0,sizeof(h));
memset(f,0,sizeof(f));
memset(fa,0,sizeof(fa));
int a,b;
for(int i=0;i<k;i++){
scanf("%d%d",&a,&b);
h[b]++;
f[a]++;
fa[a][b]++;
}
LL sum=0;
for(int i=1;i<=n;i++){
sum=0;
for(int j=1;j<=m;j++){
if(g[i][j]){
sum+=h[j];
if(fa[i][j]) sum-=fa[i][j];
}
}
if(i<n)printf("%I64d ",sum);
else printf("%I64d\n",sum);
}
}
return 0;
}