Codeforces Round #345

A Joysticks

Test instructions: About 2 of the discharge, charge 1, ask two duration of the longest is how much.

Water problem, note that when a, B is equal to 1, the direct game is over.

#include <cstdio>#include<iostream>#include<algorithm>using namespacestd;intb;intMain () { while(SCANF ("%d%d", &a,&b)! =EOF) {            intnum=0; if(a==1&&b==1) {printf ("0\n");  Break; }         while(a>0&&b>0){            if(a>b) Swap (A, a); A++;b-=2; //printf ("%d%d\n", A, b);num++; } printf ("%d\n", num); }return 0;}

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B Beautiful Paintings

Two water, is a i+1 >a i the logarithm of the maximum number, so the actual only need to ask how many of the same value, and then use the total minus this value. The reason is, since we sort, it is best to be in accordance with the order from small to large, but there may be repeated, so do not meet the conditions, so to re-shoot separately, then and the front of the broken, so as long as the total minus the number of broken.

#include <cstdio>#include<cstring>#include<iostream>using namespacestd;inta[1005],k,n;intMain () { while(SCANF ("%d", &n)! =EOF) {memset (A,0,sizeof(a)); intnum=0; intmaxx=0;  for(intI=0; i<n;i++) {scanf ("%d",&k); A[K]++; if(A[k]>maxx) maxx=A[k]; }       //printf ("%d\n", Maxx);printf"%d\n", N-Maxx); }return 0;}

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C Watchmen

Idea: Simple to xi=xj, only need to ask the situation of the yi=yj and the number of 22 combinations, and then subtract Xi=xj and Yi=yj in the case of the 22 combination mode. The map is used.

#include <cstdio>#include<cstring>#include<map>#include<iostream>using namespaceStd;typedefLong LongLl;map<pair<LL,LL>,LL>Map1;map<ll,ll>map2,map3;intMain () {intN;    LL A, B;  while(SCANF ("%d", &n)! =EOF)            {map1.clear ();            Map2.clear ();        Map3.clear ();  for(intI=0; i<n;i++) {scanf ("%i64d%i64d",&a,&b); Map1[make_pair (A, b)]++; Map2[a]++; MAP3[B]++; }        /*printf ("%d\n", Map1.size ());        printf ("%d\n", Map2.size ()); printf ("%d\n", Map3.size ());*/Map<pair<LL,LL>,LL>:: Iterator it1; LL sum=0;  for(It1=map1.begin (); It1!=map1.end (); it1++){            //if (it1->second!=1)sum+=it1->second* (it1->second-1)/2; //else sum++;} LL sum1=0; Map<LL,LL>:: Iterator it2;  for(It2=map2.begin (); It2!=map2.end (); it2++){                //if (it2->second!=1)sum1+=it2->second* (it2->second-1)/2; //else sum1++;} map<LL,LL>:: Iterator it3;  for(It3=map3.begin (); It3!=map3.end (); it3++){                //if (it3->second!=1)sum1+=it3->second* (it3->second-1)/2; //else sum1++;} printf ("%i64d\n", sum1-sum); }return 0;}

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Not to be continued ~ After the question has not written ~orz

Codeforces Round #345