uva1152 - 4 Values whose Sum is 0 入门经典II 第八章 例题8-3
2015-01-18 19:50
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题目链接:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=246&page=show_problem&problem=3593
分析:求a+b+c+d=0。分别枚举a+b,-c-d。和书上的思想一样。
#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 4000 + 10;
int a[maxn],b[maxn],c[maxn],d[maxn],ab[maxn*maxn];
int main(){
int T;
cin >> T;
while(T--){
int N;
cin >> N;
for(int i=0;i<N;i++){
cin>>a[i]>>b[i]>>c[i]>>d[i];
}
for(int i=0;i<N;i++)
for(int j=0;j<N;j++)
ab[i*N+j]=a[i]+b[j];
sort(ab,ab+N*N);
long long cnt=0;//用int也能过,不过还是要注意一下,毕竟在比赛中犯过同样的错误
for(int i=0;i<N;i++){
for(int j=0;j<N;j++){
cnt+=upper_bound(ab,ab+N*N,-c[i]-d[j]) -
lower_bound(ab,ab+N*N,-c[i]-d[j]);
}
}
printf("%lld\n",cnt);
if(T) printf("\n");
}
return 0;
}
汝佳提供的代码:
// UVa1152 4 Values Whose Sum is Zero
// Rujia Liu
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 4000 + 5;
int n, c, A[maxn], B[maxn], C[maxn], D[maxn], sums[maxn*maxn];
int main() {
int T;
scanf("%d", &T);
while(T--) {
scanf("%d", &n);
for(int i = 0; i < n; i++)
scanf("%d%d%d%d", &A[i], &B[i], &C[i], &D[i]);
c = 0;
for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++)
sums[c++] = A[i] + B[j];
sort(sums, sums+c);
long long cnt = 0;
for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++)
cnt += upper_bound(sums, sums+c, -C[i]-D[j]) - lower_bound(sums, sums+c, -C[i]-D[j]);
printf("%lld\n", cnt);
if(T) printf("\n");
}
return 0;
}
分析:求a+b+c+d=0。分别枚举a+b,-c-d。和书上的思想一样。
#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 4000 + 10;
int a[maxn],b[maxn],c[maxn],d[maxn],ab[maxn*maxn];
int main(){
int T;
cin >> T;
while(T--){
int N;
cin >> N;
for(int i=0;i<N;i++){
cin>>a[i]>>b[i]>>c[i]>>d[i];
}
for(int i=0;i<N;i++)
for(int j=0;j<N;j++)
ab[i*N+j]=a[i]+b[j];
sort(ab,ab+N*N);
long long cnt=0;//用int也能过,不过还是要注意一下,毕竟在比赛中犯过同样的错误
for(int i=0;i<N;i++){
for(int j=0;j<N;j++){
cnt+=upper_bound(ab,ab+N*N,-c[i]-d[j]) -
lower_bound(ab,ab+N*N,-c[i]-d[j]);
}
}
printf("%lld\n",cnt);
if(T) printf("\n");
}
return 0;
}
汝佳提供的代码:
// UVa1152 4 Values Whose Sum is Zero
// Rujia Liu
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 4000 + 5;
int n, c, A[maxn], B[maxn], C[maxn], D[maxn], sums[maxn*maxn];
int main() {
int T;
scanf("%d", &T);
while(T--) {
scanf("%d", &n);
for(int i = 0; i < n; i++)
scanf("%d%d%d%d", &A[i], &B[i], &C[i], &D[i]);
c = 0;
for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++)
sums[c++] = A[i] + B[j];
sort(sums, sums+c);
long long cnt = 0;
for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++)
cnt += upper_bound(sums, sums+c, -C[i]-D[j]) - lower_bound(sums, sums+c, -C[i]-D[j]);
printf("%lld\n", cnt);
if(T) printf("\n");
}
return 0;
}
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