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题解:
二维PUIQ裸题
注意3点
1.在add时,写2个for循环而不是两个while。
2.从0开始,所以全部需要+1
3.在计算sum时,画图模拟一下就知道如何做了
代码:
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #include<map> #include<set> using namespace std; using namespace std; #define pb push_back #define mp make_pair #define se second #define fs first #define ll long long #define CLR(x) memset(x,0,sizeof x) #define SZ(x) ((int)(x).size()) #define FOR(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++) typedef pair<int,int> P; const double eps=1e-9; const int maxn=100010; const int N=200001000; const int mod=9901; ll read() { ll x=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f; } //----------------------------------------------------------------------------- int m[1200][1200],c[1200][1200]; int lowbit(int x) {return x&-x;} void add(int x,int y,int v){ for(int i=x;i<1200;i+=lowbit(i)) for(int j=y;j<1200;j+=lowbit(j)) c[i][j]+=v; } ll sum(int x,int y){ ll cnt=0; for(int i=x;i;i-=lowbit(i)) for(int j=y;j;j-=lowbit(j)) cnt+=c[i][j]; return cnt; } int main(){ int x1,y1,x2,y2,val,op; while(~scanf("%d",&op)&&op!=3){ if(op==0) scanf("%d",&val); if(op==1){ scanf("%d%d%d",&x1,&y1,&val); x1++;y1++; add(x1,y1,val); } if(op==2){ scanf("%d%d%d%d",&x1,&y1,&x2,&y2); x1++;y1++;x2++;y2++; printf("%lld\n",sum(x2,y2)-sum(x1-1,y2)-sum(x2,y1-1)+sum(x1-1,y1-1)); } } return 0; }
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原文地址:http://www.cnblogs.com/byene/p/5897540.html