标签：水题
http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1585
就是旋转一下，找对应关系，注意这里枚举的顺序按原来顺序来，然后输出的时候把n和m的内外顺序换一下
#include<stdio.h>
int a[105][105],b[105][105];
int main(){
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
int T,q,n,m;
scanf("%d",&T);
while(T--){
scanf("%d%d%d",&n,&m,&q);
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
scanf("%d",&a[i][j]);
}
}
if(q==1){
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
b[j][n-i+1]=a[i][j];
}
}
for(int i=1;i<=m;i++){
for(int j=1;j<=n;j++){
//b[i][j]=a[m-j+1][i];
if(j==n) printf("%d\n",b[i][j]);
else printf("%d ",b[i][j]);
}
}
}
else{
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
b[m-j+1][i]=a[i][j];
}
}
for(int i=1;i<=m;i++){
for(int j=1;j<=n;j++){
if(j==n) printf("%d\n",b[i][j]);
else printf("%d ",b[i][j]);
}
}
}
}
}http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1588
每次优先合并最小的两个数，其实就是哈夫曼树的建法，用一个优先队列维护
#include<stdio.h>
#include<queue>
#include<iostream>
using namespace std;
int main(){
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
priority_queue<int,vector<int>,greater<int> > pq;
int T,n,tmp;
scanf("%d",&T);
while(T--){
int s=0;
scanf("%d",&n);
for(int i=0;i<n;i++) {
scanf("%d",&tmp);
pq.push(tmp);
}
int t1,t2;
for(int i=1;i<n;i++){
t1=pq.top();
pq.pop();
t2=pq.top();
pq.pop();
s+=t1+t2;
pq.push(t1+t2);
}
pq.pop();
printf("%d\n",s);
}
}
烦人的异或：
用一个数组该点到左上角的异或总值
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int s[1005][1005],l[1005][1005],h[1005][1005];
int T,n,m,q;
int main(){
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
scanf("%d",&T);
while(T--){
scanf("%d%d%d",&n,&m,&q);
memset(s,0,sizeof(s));
memset(l,0,sizeof(l));
memset(h,0,sizeof(h));
int tmp;
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
scanf("%d",&tmp);
l[i][j]=l[i][j-1]^tmp;
h[i][j]=h[i-1][j]^tmp;
s[i][j]=s[i-1][j-1]^tmp^l[i][j-1]^h[i-1][j];
}
}
for(int i=1;i<=m;i++){
//printf("%d %d\n",s[1][i],l[1][i]);
}
int x1,y1,x2,y2;
while(q--){
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
int sum=0;
sum^=s[x1-1][y1-1];
sum^=s[x1-1][m]^s[x1-1][y2];
sum^=s[x2][y2]^s[x1-1][y1-1]^(s[x1-1][y2]^s[x1-1][y1-1])^(s[x2][y1-1]^s[x1-1][y1-1]);
sum^=s[n][y1-1]^s[x2][y1-1];
sum^=s[n][m]^s[x2][y2]^(s[x2][m]^s[x2][y2])^(s[n][y2]^s[x2][y2]);
printf("%d\n",sum);
}
}
}
括号匹配：
这道题，一开始理解错了，给出的已经括号是符合运算表达式的括号，所以直接用个栈来维护就行了
#include<stack>
#include<cmath>
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define lowbit(x) -x&x
using namespace std;
const int maxn=110000;
const int INF=0x3f3f3f3f;
char str[maxn];
int match[maxn];
int tree[3][maxn];
void insert(int *a,int x){
while(x<maxn){
a[x] ++;
x += lowbit(x);
}
}
int get_sum(int *a,int x){
int res = 0;
while(x>0){
res += a[x];
x -= lowbit(x);
}
return res;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
while(~scanf("%s",str+1)){
stack<int> ss;
int s,m,l,n;
s=m=l=0;
memset(tree,0,sizeof(tree));
m=strlen(str+1);
for(int i = 1;i<=m;i++){
if(str[i]=='(' || str[i]=='[' || str[i]=='{')
ss.push(i);
else{
int t = ss.top();
ss.pop();
match[t]=i;
match[i]=t;
if(str[i]==')') insert(tree[0],t);
else if(str[i]==']') insert(tree[1],t);
else if(str[i]=='}') insert(tree[2],t);
}
}
scanf("%d",&n);
for(int i = 0;i<n;i++){
int t;
scanf("%d",&t);
int c=match[t];
//printf("%d %d\n",get_sum(tree[0],c),get_sum(tree[0],t));
printf("%d",c);
if(match[t]<t) c=t,t=match[t];
c--;
printf(" %d %d %d\n",get_sum(tree[0],c)-get_sum(tree[0],t),get_sum(tree[1],c)-get_sum(tree[1],t),get_sum(tree[2],c)-get_sum(tree[2],t));
}
printf("\n");
}
return 0;
}
标签：水题
原文地址：http://blog.csdn.net/lj94093/article/details/45443143
